Two Unknowns Become One the Moment You Add the Right Multiple
Solve a pair of linear equations by substitution and by elimination, choose the faster method for the equations in front of you, and turn word problems on ages, digits, fractions, costs, boats and shared work into two equations you can actually solve.
Why do you need algebra when the graph already gives the answer?
Because a graph can only be read as accurately as it is drawn, and most answers are not whole numbers.
Take the pair and . Its solution is and . No amount of careful plotting on graph paper will let you read those two fractions off a picture. The graphical method told you how many solutions exist; it cannot tell you what they are unless they happen to be small integers.
So this part of the chapter replaces the picture with two algebraic methods.
- Substitution — express one variable in terms of the other from one equation, and put it into the other
- Elimination — multiply the equations so that one variable has matching coefficients, then add or subtract to remove it
Both always work. Neither is always faster, and choosing well is a real skill: substitution suits an equation where a variable already has a coefficient of , while elimination suits a pair where neither variable is isolated.
And then the chapter turns to the part that carries the most marks: word problems. Ages, two-digit numbers, fractions, prices, boats in a stream, and people sharing work — all of them become two equations in two unknowns once you name the unknowns properly. The algebra is the easy half. Translating the sentences is where marks are lost, and this page spends most of its length on exactly that.
This page covers the second part of the CBSE Class 10 Maths chapter on a pair of linear equations in two variables: the substitution method, the elimination method, and situational word problems.
Take the pair and . Its solution is and . No amount of careful plotting on graph paper will let you read those two fractions off a picture. The graphical method told you how many solutions exist; it cannot tell you what they are unless they happen to be small integers.
So this part of the chapter replaces the picture with two algebraic methods.
- Substitution — express one variable in terms of the other from one equation, and put it into the other
- Elimination — multiply the equations so that one variable has matching coefficients, then add or subtract to remove it
Both always work. Neither is always faster, and choosing well is a real skill: substitution suits an equation where a variable already has a coefficient of , while elimination suits a pair where neither variable is isolated.
And then the chapter turns to the part that carries the most marks: word problems. Ages, two-digit numbers, fractions, prices, boats in a stream, and people sharing work — all of them become two equations in two unknowns once you name the unknowns properly. The algebra is the easy half. Translating the sentences is where marks are lost, and this page spends most of its length on exactly that.
This page covers the second part of the CBSE Class 10 Maths chapter on a pair of linear equations in two variables: the substitution method, the elimination method, and situational word problems.
How do you solve a pair of equations by substitution?
Make one variable the subject of one equation, put that expression into the other equation, solve the single-variable equation you get, and then go back for the second variable.
The four steps, which should be visible in your answer.
- Step 1 — pick the equation where a variable is easiest to isolate, and isolate it
- Step 2 — substitute that expression into the other equation
- Step 3 — solve the resulting equation in one variable
- Step 4 — substitute the value back to find the second variable, then check in both original equations
Worked example 1. Solve and .
The second equation has with a coefficient of , so isolate there:
Substitute into the first:
Then
Check in the first equation:
Correct — and notice that no graph could have produced those fractions.
Worked example 2 — the trap to watch for. Solve and by substitution.
From the second, , so . Substituting into the first:
and . **The solution is , .
Check both**: and . Both satisfied.
The single mistake that wastes a whole question. Substituting back into the same equation you rearranged. Putting into gives , which is — true but useless. It carries no information, because you only rewrote the equation you started from. The substitution must go into the other equation, and a line that collapses to is a signal that you went back to the wrong one.
When substitution is the right choice. When some variable already has a coefficient of or , so isolating it introduces no fractions. In that is obvious; in it is not, and isolating there would leave you dividing by throughout.
The four steps, which should be visible in your answer.
- Step 1 — pick the equation where a variable is easiest to isolate, and isolate it
- Step 2 — substitute that expression into the other equation
- Step 3 — solve the resulting equation in one variable
- Step 4 — substitute the value back to find the second variable, then check in both original equations
Worked example 1. Solve and .
The second equation has with a coefficient of , so isolate there:
Substitute into the first:
Then
Check in the first equation:
Correct — and notice that no graph could have produced those fractions.
Worked example 2 — the trap to watch for. Solve and by substitution.
From the second, , so . Substituting into the first:
and . **The solution is , .
Check both**: and . Both satisfied.
The single mistake that wastes a whole question. Substituting back into the same equation you rearranged. Putting into gives , which is — true but useless. It carries no information, because you only rewrote the equation you started from. The substitution must go into the other equation, and a line that collapses to is a signal that you went back to the wrong one.
When substitution is the right choice. When some variable already has a coefficient of or , so isolating it introduces no fractions. In that is obvious; in it is not, and isolating there would leave you dividing by throughout.
When is elimination faster, and how do you set it up?
When neither variable has a coefficient of one. Multiply each equation by a suitable number so that one variable has the same coefficient in both, then add or subtract to knock it out.
The four steps.
- Step 1 — decide which variable to eliminate, usually the one with the smaller or more convenient coefficients
- Step 2 — multiply one or both equations so that that variable's coefficients match in size
- Step 3 — add if the matched coefficients have opposite signs, subtract if they have the same sign
- Step 4 — solve for the remaining variable, back-substitute, and check in both originals
Worked example 1. Solve and by elimination.
The coefficients are and . Multiplying the second equation by makes them and :
Adding, since the signs are opposite:
Substituting into :
**The solution is , — the same answer substitution gave, as it must.
Worked example 2 — matching both ways.** Solve and .
To eliminate , multiply the second equation by so that its coefficient becomes :
Subtracting, since the signs now match:
Then , so and .
Check: . Correct.
Worked example 3 — eliminating the other variable as a check. Take the same pair and eliminate instead. Multiply the first by and the second by :
Subtracting: , so — the same value, reached by a different route. That is a genuine independent check, and it is worth doing when a question is worth several marks.
Add or subtract? The rule stated plainly. After matching the coefficients, look at the signs of the matched terms. Opposite signs mean add; identical signs mean subtract. Getting this backwards produces a term that does not vanish, and the give-away is that you are still looking at two variables after the step that was meant to remove one.
What each method reveals when there is no unique solution. If the pair is dependent, both methods collapse to a statement like , which is true and tells you there are infinitely many solutions. If the pair is inconsistent, they collapse to something false like , which tells you there is no solution. The algebra reports the same three cases the ratio test predicted, so the two halves of the chapter agree.
The four steps.
- Step 1 — decide which variable to eliminate, usually the one with the smaller or more convenient coefficients
- Step 2 — multiply one or both equations so that that variable's coefficients match in size
- Step 3 — add if the matched coefficients have opposite signs, subtract if they have the same sign
- Step 4 — solve for the remaining variable, back-substitute, and check in both originals
Worked example 1. Solve and by elimination.
The coefficients are and . Multiplying the second equation by makes them and :
Adding, since the signs are opposite:
Substituting into :
**The solution is , — the same answer substitution gave, as it must.
Worked example 2 — matching both ways.** Solve and .
To eliminate , multiply the second equation by so that its coefficient becomes :
Subtracting, since the signs now match:
Then , so and .
Check: . Correct.
Worked example 3 — eliminating the other variable as a check. Take the same pair and eliminate instead. Multiply the first by and the second by :
Subtracting: , so — the same value, reached by a different route. That is a genuine independent check, and it is worth doing when a question is worth several marks.
Add or subtract? The rule stated plainly. After matching the coefficients, look at the signs of the matched terms. Opposite signs mean add; identical signs mean subtract. Getting this backwards produces a term that does not vanish, and the give-away is that you are still looking at two variables after the step that was meant to remove one.
What each method reveals when there is no unique solution. If the pair is dependent, both methods collapse to a statement like , which is true and tells you there are infinitely many solutions. If the pair is inconsistent, they collapse to something false like , which tells you there is no solution. The algebra reports the same three cases the ratio test predicted, so the two halves of the chapter agree.
How do you turn a word problem into two equations?
Name the two unknowns with letters in the first line, translate each sentence of the problem into one equation, and only then start solving. Most lost marks in this chapter are lost before any algebra begins.
Worked example 1 — ages. A father is three times as old as his son. After years he will be twice as old as his son. Find their present ages.
Let the father's present age be years and the son's be years.
- "Three times as old" gives
- "After years" adds to both ages, giving
Substituting into the second:
so . **The father is and the son is .
Check both conditions**: , and after years the ages are and , where . Both hold.
The trap here is adding the years to only one age. Twelve years pass for everybody, so both expressions must change.
Worked example 2 — a two-digit number. The sum of the digits of a two-digit number is . Nine times the number equals twice the number obtained by reversing its digits. Find the number.
Let the tens digit be and the units digit be . Then
- **The number itself is **, not
- **The reversed number is **
The two conditions give
Expanding the second: , so and therefore . Substituting into the first:
**The number is .
Check**: , the reversed number is , and . Correct.
The place-value step is the whole difficulty. A two-digit number with digits and is , and writing it as turns a linear problem into a nonsense one.
Worked example 3 — a fraction. A fraction becomes when is added to both its numerator and its denominator, and becomes when is added to both. Find the fraction.
Let the fraction be . Then
Eliminating : multiply the first by and the second by :
Subtracting gives . Then , so and .
**The fraction is .
Check**: from , and from . Both conditions are met.
Worked example 4 — cost of items. Five pens and six pencils cost ₹; three pens and two pencils cost ₹. Find the cost of each.
This is the pair solved in the previous section: **a pen costs ₹ and a pencil costs ₹.** Notice that the translation was immediate — each sentence gave one equation with no rearranging at all, which is why cost problems are the easiest family in this chapter and a good place to start revising.
Worked example 1 — ages. A father is three times as old as his son. After years he will be twice as old as his son. Find their present ages.
Let the father's present age be years and the son's be years.
- "Three times as old" gives
- "After years" adds to both ages, giving
Substituting into the second:
so . **The father is and the son is .
Check both conditions**: , and after years the ages are and , where . Both hold.
The trap here is adding the years to only one age. Twelve years pass for everybody, so both expressions must change.
Worked example 2 — a two-digit number. The sum of the digits of a two-digit number is . Nine times the number equals twice the number obtained by reversing its digits. Find the number.
Let the tens digit be and the units digit be . Then
- **The number itself is **, not
- **The reversed number is **
The two conditions give
Expanding the second: , so and therefore . Substituting into the first:
**The number is .
Check**: , the reversed number is , and . Correct.
The place-value step is the whole difficulty. A two-digit number with digits and is , and writing it as turns a linear problem into a nonsense one.
Worked example 3 — a fraction. A fraction becomes when is added to both its numerator and its denominator, and becomes when is added to both. Find the fraction.
Let the fraction be . Then
Eliminating : multiply the first by and the second by :
Subtracting gives . Then , so and .
**The fraction is .
Check**: from , and from . Both conditions are met.
Worked example 4 — cost of items. Five pens and six pencils cost ₹; three pens and two pencils cost ₹. Find the cost of each.
This is the pair solved in the previous section: **a pen costs ₹ and a pencil costs ₹.** Notice that the translation was immediate — each sentence gave one equation with no rearranging at all, which is why cost problems are the easiest family in this chapter and a good place to start revising.
How do speed, boat and shared-work problems become linear equations?
By writing the two given situations as two separate equations in the same two unknowns. For motion the unknowns are usually a speed and a time; for boats they are the speed in still water and the speed of the stream; for work they are the one-day work of each worker.
Worked example 1 — a boat in a stream. A boat covers km downstream in hours and the same km upstream in hours. Find the speed of the boat in still water and the speed of the stream.
Let the boat's speed in still water be km/h and the stream's speed be km/h. Then
- Downstream the stream helps, so the effective speed is
- Upstream the stream opposes, so the effective speed is
From the two journeys,
Adding: , so . Subtracting: , so .
**The boat's speed in still water is km/h and the stream's speed is km/h.
Check**: and . Both agree with the journeys.
Worked example 2 — a train's distance. A train travels a certain distance at a uniform speed. Had the speed been km/h more, it would have taken hours less; had it been km/h less, it would have taken hours more. Find the distance.
Let the speed be km/h and the time be hours, so the distance is km. The distance is the same in all three descriptions, which is what generates the equations:
From the first, . Substituting:
and . **The distance is km.
Check both alternatives**: at km/h the time is hours, which is hours less; at km/h it is hours, which is hours more. Both conditions verified.
Notice what made this work. The terms cancelled on both sides, leaving two linear equations. If they had not cancelled the problem would have been quadratic, and that is exactly what happens in the next chapter — so the cancellation is worth watching for.
Worked example 3 — shared work. Two men and five boys can finish a piece of work in days, while three men and six boys can finish the same work in days. Find the time one man alone and one boy alone would take.
Never take the unknowns to be the days. Take them to be the work done in one day, because that is what adds up. Let one man do of the work per day and one boy do :
Multiply the first by and the second by :
Subtracting: , so . Then , giving and .
**One man alone would take days and one boy alone days.
Check**: , which is one quarter of the work per day and so days in all. Correct, and the second condition checks the same way: , giving days.
The single idea behind every work problem. Rates add; times do not. A man who takes days and a boy who takes days do not together take days — they take , because .
Worked example 1 — a boat in a stream. A boat covers km downstream in hours and the same km upstream in hours. Find the speed of the boat in still water and the speed of the stream.
Let the boat's speed in still water be km/h and the stream's speed be km/h. Then
- Downstream the stream helps, so the effective speed is
- Upstream the stream opposes, so the effective speed is
From the two journeys,
Adding: , so . Subtracting: , so .
**The boat's speed in still water is km/h and the stream's speed is km/h.
Check**: and . Both agree with the journeys.
Worked example 2 — a train's distance. A train travels a certain distance at a uniform speed. Had the speed been km/h more, it would have taken hours less; had it been km/h less, it would have taken hours more. Find the distance.
Let the speed be km/h and the time be hours, so the distance is km. The distance is the same in all three descriptions, which is what generates the equations:
From the first, . Substituting:
and . **The distance is km.
Check both alternatives**: at km/h the time is hours, which is hours less; at km/h it is hours, which is hours more. Both conditions verified.
Notice what made this work. The terms cancelled on both sides, leaving two linear equations. If they had not cancelled the problem would have been quadratic, and that is exactly what happens in the next chapter — so the cancellation is worth watching for.
Worked example 3 — shared work. Two men and five boys can finish a piece of work in days, while three men and six boys can finish the same work in days. Find the time one man alone and one boy alone would take.
Never take the unknowns to be the days. Take them to be the work done in one day, because that is what adds up. Let one man do of the work per day and one boy do :
Multiply the first by and the second by :
Subtracting: , so . Then , giving and .
**One man alone would take days and one boy alone days.
Check**: , which is one quarter of the work per day and so days in all. Correct, and the second condition checks the same way: , giving days.
The single idea behind every work problem. Rates add; times do not. A man who takes days and a boy who takes days do not together take days — they take , because .
Exam tip
Which habits protect the marks in a word problem?
Declare the unknowns with their units in the first line, and finish by answering the question that was asked. An examiner marks the translation, the solving and the interpretation as three separate things.
- **Write "let the father's age be years"** — with the unit. A bare "let " loses the framing mark
- Translate one sentence at a time and write each equation on its own line before combining anything
- **A two-digit number with digits and is **, and the reversed number is
- Add the same number of years to every age in an "after years" condition
- In work problems take the one-day work as the unknown, never the number of days
- **For boats, downstream is and upstream is — write both before touching the numbers
- Choose the method deliberately**: substitution when a coefficient is , elimination otherwise
- After matching coefficients, add if the signs differ and subtract if they agree
- Check in both original equations, not in the rearranged one
- Answer in words at the end — "the fraction is " — and give units where the quantity has them
The misconception to name. Substituting your rearranged expression back into the equation you rearranged is not a check and not a step. **It always produces a true statement such as , because you have only rewritten one equation twice. The information in a pair lives in the fact that the two equations are different, and using one of them twice throws half the problem away.
A second trap. Solving correctly and then not answering the question. A problem that asks for a number** wants , not " and "; a problem that asks for a distance wants km, not the speed and the time you found on the way. The last line of the answer should repeat the question's own words.
- **Write "let the father's age be years"** — with the unit. A bare "let " loses the framing mark
- Translate one sentence at a time and write each equation on its own line before combining anything
- **A two-digit number with digits and is **, and the reversed number is
- Add the same number of years to every age in an "after years" condition
- In work problems take the one-day work as the unknown, never the number of days
- **For boats, downstream is and upstream is — write both before touching the numbers
- Choose the method deliberately**: substitution when a coefficient is , elimination otherwise
- After matching coefficients, add if the signs differ and subtract if they agree
- Check in both original equations, not in the rearranged one
- Answer in words at the end — "the fraction is " — and give units where the quantity has them
The misconception to name. Substituting your rearranged expression back into the equation you rearranged is not a check and not a step. **It always produces a true statement such as , because you have only rewritten one equation twice. The information in a pair lives in the fact that the two equations are different, and using one of them twice throws half the problem away.
A second trap. Solving correctly and then not answering the question. A problem that asks for a number** wants , not " and "; a problem that asks for a distance wants km, not the speed and the time you found on the way. The last line of the answer should repeat the question's own words.
Did you know
Why does the same pair of equations solve a boat and a bicycle?
Look at the boat problem again, stripped of its story. Two quantities combine by addition in one situation and by subtraction in the other, and you are told the result in each case. That is all the mathematics knows about it.
Which is why the identical pair of equations describes a surprising number of different situations:
- A boat and a stream — downstream , upstream
- An aeroplane and a wind — with the wind , against it
- A cyclist on a slope — downhill and uphill, in the simplest model
- A man rowing and walking, where the two legs of a journey combine differently
In every case the algebra is the same and only the nouns change. That is not a coincidence about boats. It is what algebra is for — the letters hold a shape of reasoning that can be re-used on any situation with the same shape.
The same thing happens with the number problems. A two-digit number problem and a coin problem look nothing alike:
- A number with digits and has value
- A purse with ten-rupee notes and one-rupee coins holds rupees
Identical expression, identical algebra. Once you see that a two-digit number is a purse of tens and units, the place-value step stops feeling like a rule to remember and becomes something obvious.
And it explains why word problems are set at all. The arithmetic in these questions is easy — a pair of linear equations takes four lines. What is being tested is whether you can recognise a familiar shape inside an unfamiliar story, which is the part of mathematics that transfers to everything else.
One last observation about the work problems. The reason rates add while times do not is the same reason speeds add in the boat problem: rates are quantities per unit of something, and per-unit quantities are the ones that add. Two taps filling a tank, two men doing a job, a boat and a stream — all of them add rates. A student who reaches for "average the times" in a work problem has spotted the wrong shape, and the check that catches it takes one line: a man who takes days and a boy who takes days finish together in days, which is faster than either alone, exactly as it must be.
Which is why the identical pair of equations describes a surprising number of different situations:
- A boat and a stream — downstream , upstream
- An aeroplane and a wind — with the wind , against it
- A cyclist on a slope — downhill and uphill, in the simplest model
- A man rowing and walking, where the two legs of a journey combine differently
In every case the algebra is the same and only the nouns change. That is not a coincidence about boats. It is what algebra is for — the letters hold a shape of reasoning that can be re-used on any situation with the same shape.
The same thing happens with the number problems. A two-digit number problem and a coin problem look nothing alike:
- A number with digits and has value
- A purse with ten-rupee notes and one-rupee coins holds rupees
Identical expression, identical algebra. Once you see that a two-digit number is a purse of tens and units, the place-value step stops feeling like a rule to remember and becomes something obvious.
And it explains why word problems are set at all. The arithmetic in these questions is easy — a pair of linear equations takes four lines. What is being tested is whether you can recognise a familiar shape inside an unfamiliar story, which is the part of mathematics that transfers to everything else.
One last observation about the work problems. The reason rates add while times do not is the same reason speeds add in the boat problem: rates are quantities per unit of something, and per-unit quantities are the ones that add. Two taps filling a tank, two men doing a job, a boat and a stream — all of them add rates. A student who reaches for "average the times" in a work problem has spotted the wrong shape, and the check that catches it takes one line: a man who takes days and a boy who takes days finish together in days, which is faster than either alone, exactly as it must be.
Exam relevance
How do these methods matter for JEE?
This is foundation work for Class 11 Straight Lines and Class 12 Matrices and Determinants, and for the word-problem and system-solving questions in JEE Main.
Where elimination leads. Class 12 systematises it into Gaussian elimination and the matrix method, and the same operations you perform by hand here — multiply an equation by a constant, add one equation to another — become the elementary row operations used to solve three equations in three unknowns and to find the inverse of a matrix. Every step you take in elimination is one of those operations, so the Class 12 procedure is not a new idea but the same idea written in a table.
Where substitution leads. It remains the workhorse for any system where one equation is easy to rearrange, including the non-linear systems that appear in Coordinate Geometry — finding where a line meets a circle or a parabola is substitution followed by a quadratic. JEE Main sets that shape constantly, and the habit of substituting into the other equation is exactly what it needs.
Where the word problems lead. Rate problems reappear throughout: relative velocity in Kinematics, work and time in arithmetic reasoning, and mixture problems in Solutions in Chemistry. The principle that rates add is the same one behind relative velocity, and a student comfortable with the boat problem has already met the reasoning.
Where the dependent and inconsistent cases lead. Class 12 measures them with the rank of the coefficient matrix and of the augmented matrix, and the collapse to or that you see here is precisely what a rank comparison detects. JEE Main asks for the value of a parameter that makes a system inconsistent, which is this chapter's find- question in matrix clothing.
Question types to expect. At this level: solve by substitution, solve by elimination, and a full word problem carrying several marks. In competitive papers: systems with a parameter, line-and-curve intersections, and reasoning items on when a system fails to have a unique solution.
The single trap that costs marks. Substituting back into the equation you rearranged. It yields a tautology and no information, and in a three-equation Class 12 system the same error shows up as a row of zeros that a student then misreads as a genuine result.
A second trap. Taking the number of days as the unknown in a work problem. Days do not add; one-day work does — and the same mistake in Class 11 Physics becomes averaging speeds instead of averaging over time, which is the classic average-speed error.
Board versus competitive emphasis. The CBSE paper marks the declaration of variables, both equations, the solving and the final sentence; a competitive paper marks only the number. The transferable habit is naming the unknowns as the quantities that add — rates, not times; place values, not digits side by side — because that single choice decides whether the equations come out linear or unusable.
Where elimination leads. Class 12 systematises it into Gaussian elimination and the matrix method, and the same operations you perform by hand here — multiply an equation by a constant, add one equation to another — become the elementary row operations used to solve three equations in three unknowns and to find the inverse of a matrix. Every step you take in elimination is one of those operations, so the Class 12 procedure is not a new idea but the same idea written in a table.
Where substitution leads. It remains the workhorse for any system where one equation is easy to rearrange, including the non-linear systems that appear in Coordinate Geometry — finding where a line meets a circle or a parabola is substitution followed by a quadratic. JEE Main sets that shape constantly, and the habit of substituting into the other equation is exactly what it needs.
Where the word problems lead. Rate problems reappear throughout: relative velocity in Kinematics, work and time in arithmetic reasoning, and mixture problems in Solutions in Chemistry. The principle that rates add is the same one behind relative velocity, and a student comfortable with the boat problem has already met the reasoning.
Where the dependent and inconsistent cases lead. Class 12 measures them with the rank of the coefficient matrix and of the augmented matrix, and the collapse to or that you see here is precisely what a rank comparison detects. JEE Main asks for the value of a parameter that makes a system inconsistent, which is this chapter's find- question in matrix clothing.
Question types to expect. At this level: solve by substitution, solve by elimination, and a full word problem carrying several marks. In competitive papers: systems with a parameter, line-and-curve intersections, and reasoning items on when a system fails to have a unique solution.
The single trap that costs marks. Substituting back into the equation you rearranged. It yields a tautology and no information, and in a three-equation Class 12 system the same error shows up as a row of zeros that a student then misreads as a genuine result.
A second trap. Taking the number of days as the unknown in a work problem. Days do not add; one-day work does — and the same mistake in Class 11 Physics becomes averaging speeds instead of averaging over time, which is the classic average-speed error.
Board versus competitive emphasis. The CBSE paper marks the declaration of variables, both equations, the solving and the final sentence; a competitive paper marks only the number. The transferable habit is naming the unknowns as the quantities that add — rates, not times; place values, not digits side by side — because that single choice decides whether the equations come out linear or unusable.
Key takeaways
What must you be able to do from this part?
Two methods, one translation habit and one check.
- Substitution: isolate a variable in one equation, put it into the other, solve, back-substitute, check
- **Use substitution when a variable already has a coefficient of **, as does in
- Elimination: match one variable's coefficients, then add if the signs differ and subtract if they agree
- Eliminating the other variable instead is a genuine independent check on a long question
- **A dependent pair collapses to and an inconsistent pair to something false, matching the ratio test of Part 1
- with ** gives , — fractions a graph could never give
- ** with ** gives by either method
- Ages: a father three times his son's age who will be twice it after years is with a son of
- **A two-digit number is ** and its reverse is — the number with digit sum satisfying the reversal condition is
- Fractions: adding to both parts giving and adding giving identifies
- Costs: five pens and six pencils at ₹ with three pens and two pencils at ₹ gives ₹ and ₹
- Boats: downstream is and upstream is , so km in hours and hours gives km/h and km/h
- Uniform motion: the same distance written three ways gives km at km/h in hours
- Work: take the one-day work as the unknown, so days and days separately means days together
- Answer the question that was asked, in words and with units
The best self-test is one problem solved twice. Take the pens-and-pencils pair, solve it by substitution and then again by elimination, and see whether both routes land on ₹ and ₹ without a single line of arithmetic in common.
- Substitution: isolate a variable in one equation, put it into the other, solve, back-substitute, check
- **Use substitution when a variable already has a coefficient of **, as does in
- Elimination: match one variable's coefficients, then add if the signs differ and subtract if they agree
- Eliminating the other variable instead is a genuine independent check on a long question
- **A dependent pair collapses to and an inconsistent pair to something false, matching the ratio test of Part 1
- with ** gives , — fractions a graph could never give
- ** with ** gives by either method
- Ages: a father three times his son's age who will be twice it after years is with a son of
- **A two-digit number is ** and its reverse is — the number with digit sum satisfying the reversal condition is
- Fractions: adding to both parts giving and adding giving identifies
- Costs: five pens and six pencils at ₹ with three pens and two pencils at ₹ gives ₹ and ₹
- Boats: downstream is and upstream is , so km in hours and hours gives km/h and km/h
- Uniform motion: the same distance written three ways gives km at km/h in hours
- Work: take the one-day work as the unknown, so days and days separately means days together
- Answer the question that was asked, in words and with units
The best self-test is one problem solved twice. Take the pens-and-pencils pair, solve it by substitution and then again by elimination, and see whether both routes land on ₹ and ₹ without a single line of arithmetic in common.