Two Equations Pin Down Two Unknowns and One Equation Cannot
Solve a pair of linear equations by substitution and by elimination, clear fractions and brackets before you start, and check a given solution or find a missing constant from it.
Why can one equation never fix two unknowns?
Take the single equation and try to find and .
You cannot. works. So does , and , and . There are infinitely many answers, and nothing in the equation chooses between them.
Drawn on a graph, those solutions form a straight line — every point on it satisfies the equation.
Now add a second equation, . That one has its own infinite family of solutions, and its own straight line.
The pair is satisfied only by a point lying on both lines — and two straight lines that are not parallel meet at exactly one point. That point is , , and it is the only answer.
So a pair of equations narrows infinitely many possibilities down to one, and it needs both of them to do it. That is why they are called simultaneous — they must hold at the same time, and a value satisfying only one of them is not a solution at all.
The rest of this page is two methods for finding that point without drawing anything.
- Substitution — make one equation give you one letter in terms of the other, and put it into the second equation
- Elimination — scale the equations so that one letter cancels when you add or subtract them
This page covers the first part of the ICSE Class 9 Mathematics chapter on simultaneous linear equations: substitution, elimination, equations that must first be cleared of fractions or brackets, and verifying a solution or finding an unknown constant from it.
You cannot. works. So does , and , and . There are infinitely many answers, and nothing in the equation chooses between them.
Drawn on a graph, those solutions form a straight line — every point on it satisfies the equation.
Now add a second equation, . That one has its own infinite family of solutions, and its own straight line.
The pair is satisfied only by a point lying on both lines — and two straight lines that are not parallel meet at exactly one point. That point is , , and it is the only answer.
So a pair of equations narrows infinitely many possibilities down to one, and it needs both of them to do it. That is why they are called simultaneous — they must hold at the same time, and a value satisfying only one of them is not a solution at all.
The rest of this page is two methods for finding that point without drawing anything.
- Substitution — make one equation give you one letter in terms of the other, and put it into the second equation
- Elimination — scale the equations so that one letter cancels when you add or subtract them
This page covers the first part of the ICSE Class 9 Mathematics chapter on simultaneous linear equations: substitution, elimination, equations that must first be cleared of fractions or brackets, and verifying a solution or finding an unknown constant from it.
How does the substitution method work?
Rearrange whichever equation is simplest to get one letter alone, then put that expression into the other equation.
The point of substitution is to turn two equations in two unknowns into one equation in one unknown, which you already know how to solve.
Worked example. Solve
Step 1 — choose the easier equation and make one letter the subject. Equation (2) has a coefficient of on , so it rearranges with no fractions:
Step 2 — substitute into the other equation. Put into (1):
Step 3 — put the value back to find the other letter. From (3):
Step 4 — check in BOTH original equations.
- In (1): . Correct
- In (2): . Correct
So the solution is , .
Now the judgement that makes substitution quick or slow. You may rearrange either equation for either letter, and the four choices are not equally pleasant.
Here, rearranging (1) for would have given — correct, but it carries a fraction into every line that follows.
**So look for a coefficient of or and make that letter the subject.** If there is one, substitution is the faster method; if every coefficient is or more, elimination usually wins — and that is the whole basis for choosing between the two methods.
One warning about the substitution itself. Put the expression in brackets before multiplying. Writing instead of loses the from the second term and is the commonest error in this method. The bracket is not optional.
The point of substitution is to turn two equations in two unknowns into one equation in one unknown, which you already know how to solve.
Worked example. Solve
Step 1 — choose the easier equation and make one letter the subject. Equation (2) has a coefficient of on , so it rearranges with no fractions:
Step 2 — substitute into the other equation. Put into (1):
Step 3 — put the value back to find the other letter. From (3):
Step 4 — check in BOTH original equations.
- In (1): . Correct
- In (2): . Correct
So the solution is , .
Now the judgement that makes substitution quick or slow. You may rearrange either equation for either letter, and the four choices are not equally pleasant.
Here, rearranging (1) for would have given — correct, but it carries a fraction into every line that follows.
**So look for a coefficient of or and make that letter the subject.** If there is one, substitution is the faster method; if every coefficient is or more, elimination usually wins — and that is the whole basis for choosing between the two methods.
One warning about the substitution itself. Put the expression in brackets before multiplying. Writing instead of loses the from the second term and is the commonest error in this method. The bracket is not optional.
How does the elimination method work?
Multiply one or both equations by numbers that make the coefficients of one letter match, then add or subtract to remove it.
The rule for the last step is worth stating plainly: if the matched coefficients have the same sign, subtract; if they have opposite signs, add.
Worked example 1 — only one equation needs multiplying. Solve
The coefficients are and . Multiplying (2) by makes them and :
The signs are opposite, so add (1) and (3):
Substitute into (2): , so and .
Check in both: and , as required. So , .
Worked example 2 — both equations need multiplying. Solve
To eliminate , the coefficients and must be made equal in size. The LCM of and is , so multiply (1) by and (2) by :
Opposite signs, so add:
Substitute into (2): , so , giving and .
Check in both: and , as required. So , .
Two points of technique worth fixing.
Multiply EVERY term, including the right-hand side. In example 2, multiplying (1) by turns into — forgetting the right-hand side is the commonest error in elimination, and it produces an answer that fails the check.
Choose the letter that is cheaper to eliminate. In example 2, eliminating would have needed the LCM of and , which is — larger numbers for the same work. Look at both letters and pick the one with the smaller LCM.
And a useful observation about the two methods. Elimination and substitution always give the same answer, because they are two routes to the same intersection point. So if a question does not specify a method, use the one that avoids fractions — substitution when some coefficient is , elimination otherwise.
The rule for the last step is worth stating plainly: if the matched coefficients have the same sign, subtract; if they have opposite signs, add.
Worked example 1 — only one equation needs multiplying. Solve
The coefficients are and . Multiplying (2) by makes them and :
The signs are opposite, so add (1) and (3):
Substitute into (2): , so and .
Check in both: and , as required. So , .
Worked example 2 — both equations need multiplying. Solve
To eliminate , the coefficients and must be made equal in size. The LCM of and is , so multiply (1) by and (2) by :
Opposite signs, so add:
Substitute into (2): , so , giving and .
Check in both: and , as required. So , .
Two points of technique worth fixing.
Multiply EVERY term, including the right-hand side. In example 2, multiplying (1) by turns into — forgetting the right-hand side is the commonest error in elimination, and it produces an answer that fails the check.
Choose the letter that is cheaper to eliminate. In example 2, eliminating would have needed the LCM of and , which is — larger numbers for the same work. Look at both letters and pick the one with the smaller LCM.
And a useful observation about the two methods. Elimination and substitution always give the same answer, because they are two routes to the same intersection point. So if a question does not specify a method, use the one that avoids fractions — substitution when some coefficient is , elimination otherwise.
How do you handle equations with fractions or brackets?
**Clear the fractions or expand the brackets first, so that the pair is in the standard form before you choose a method.
Neither method works reliably on an equation that is not yet in standard form, so this is a genuine first step and not a tidying-up option.
Worked example 1 — fractions.** Solve
Multiply each equation through by the LCM of its denominators, which is for both:
Now eliminate . The LCM of and is , so multiply (3) by and (4) by :
The signs are the same, so subtract:
Substitute into (3): , so , giving and .
Check in the ORIGINAL equations, not the cleared ones.
- In (1): , as required
- In (2): , as required
So , .
Worked example 2 — brackets. Solve
Expand and collect. From (1):
From (2):
Now eliminate : multiply (3) by and (4) by :
Opposite signs, so add: , giving . Then from (3): , so .
Check in the original bracketed equations.
- In (1): , as required
- In (2): , as required
So , .
Notice the sign trap in example 2. Expanding gives , not . A minus outside a bracket changes every sign inside it, and getting that wrong shifts the constant and produces a wrong pair of equations that will still solve neatly — which is why it goes undetected without a check.
And notice where the check was done. Substituting into the original equations, not into (3) and (4), is what tests the clearing step as well as the solving. If you check only in the cleared equations, an error made while clearing them will never show up.
Neither method works reliably on an equation that is not yet in standard form, so this is a genuine first step and not a tidying-up option.
Worked example 1 — fractions.** Solve
Multiply each equation through by the LCM of its denominators, which is for both:
Now eliminate . The LCM of and is , so multiply (3) by and (4) by :
The signs are the same, so subtract:
Substitute into (3): , so , giving and .
Check in the ORIGINAL equations, not the cleared ones.
- In (1): , as required
- In (2): , as required
So , .
Worked example 2 — brackets. Solve
Expand and collect. From (1):
From (2):
Now eliminate : multiply (3) by and (4) by :
Opposite signs, so add: , giving . Then from (3): , so .
Check in the original bracketed equations.
- In (1): , as required
- In (2): , as required
So , .
Notice the sign trap in example 2. Expanding gives , not . A minus outside a bracket changes every sign inside it, and getting that wrong shifts the constant and produces a wrong pair of equations that will still solve neatly — which is why it goes undetected without a check.
And notice where the check was done. Substituting into the original equations, not into (3) and (4), is what tests the clearing step as well as the solving. If you check only in the cleared equations, an error made while clearing them will never show up.
How do you verify a solution or find a missing constant?
Substitute the given values into every equation — and if a constant is unknown, that substitution becomes an equation for the constant.
Verifying a solution. A pair of values is a solution only if it satisfies both equations. Checking one is not verification.
Worked example 1. Is , a solution of and ?
- First: , as required
- Second: , as required
Both hold, so yes, it is a solution.
Worked example 2. Is , a solution of the same pair?
- First: , as required
- Second: , which is not , so this pair fails
So the answer is no, even though the first equation was satisfied perfectly. A value that fits one equation is on one line and not on the other, so it is not the intersection point.
Finding one unknown constant. If a solution is given, substituting it turns the unknown constant into the only thing left to find.
Worked example 3. If , satisfies , find .
Check: , as required
Finding two unknown constants. With two unknown constants you need two equations — and substituting the given solution into both supplies them.
Worked example 4. If , is the solution of
find and .
Substitute , into both:
**These are now simultaneous equations in and .** From (1), . Substitute into (2):
Then .
Check in both original equations with , , , :
- , as required
- , as required
So and .
Now the thing worth noticing about example 4, because it is what the question is really testing. The unknowns swapped roles completely. You were given and and asked for and — so the letters that are usually constants became the variables, and the letters that are usually variables became known numbers.
Nothing in the method changed. A pair of linear equations in two unknowns is solved the same way whatever the unknowns are called.
So the real skill in this section is reading which letters are unknown, and a question of this shape is checking that you can substitute in the direction the question requires rather than the direction you are used to.
Verifying a solution. A pair of values is a solution only if it satisfies both equations. Checking one is not verification.
Worked example 1. Is , a solution of and ?
- First: , as required
- Second: , as required
Both hold, so yes, it is a solution.
Worked example 2. Is , a solution of the same pair?
- First: , as required
- Second: , which is not , so this pair fails
So the answer is no, even though the first equation was satisfied perfectly. A value that fits one equation is on one line and not on the other, so it is not the intersection point.
Finding one unknown constant. If a solution is given, substituting it turns the unknown constant into the only thing left to find.
Worked example 3. If , satisfies , find .
Check: , as required
Finding two unknown constants. With two unknown constants you need two equations — and substituting the given solution into both supplies them.
Worked example 4. If , is the solution of
find and .
Substitute , into both:
**These are now simultaneous equations in and .** From (1), . Substitute into (2):
Then .
Check in both original equations with , , , :
- , as required
- , as required
So and .
Now the thing worth noticing about example 4, because it is what the question is really testing. The unknowns swapped roles completely. You were given and and asked for and — so the letters that are usually constants became the variables, and the letters that are usually variables became known numbers.
Nothing in the method changed. A pair of linear equations in two unknowns is solved the same way whatever the unknowns are called.
So the real skill in this section is reading which letters are unknown, and a question of this shape is checking that you can substitute in the direction the question requires rather than the direction you are used to.
Exam tip
Exam tip: number your equations and check in both originals
Number every equation as , , and refer to the numbers in your working. Examiners follow the reasoning through the numbers, and method marks depend on it.
**Get the pair into the form first — clear fractions by multiplying through by the LCM of the denominators, and expand all brackets.
A minus outside a bracket changes EVERY sign inside it.** .
**For substitution, make the subject a letter whose coefficient is or . If there is none, prefer elimination.
Put the substituted expression in BRACKETS** — , never .
For elimination, multiply EVERY term including the right-hand side.
Same signs — subtract. Opposite signs — add. Say which you are doing.
Choose the letter with the smaller LCM to eliminate; it keeps the numbers small.
Always find the second unknown by substituting into the SIMPLER original equation, not into one you have multiplied.
Check in BOTH original equations — and for a question involving fractions or brackets, check in the originals rather than the cleared versions, so that the clearing step is tested too.
Verifying means both equations. A pair satisfying only one is not a solution.
To find a constant, substitute the given solution — with one constant you get one equation, and with two constants you get a pair to solve.
And state the answer as a pair, and , rather than leaving two isolated numbers at the foot of the page.
**Get the pair into the form first — clear fractions by multiplying through by the LCM of the denominators, and expand all brackets.
A minus outside a bracket changes EVERY sign inside it.** .
**For substitution, make the subject a letter whose coefficient is or . If there is none, prefer elimination.
Put the substituted expression in BRACKETS** — , never .
For elimination, multiply EVERY term including the right-hand side.
Same signs — subtract. Opposite signs — add. Say which you are doing.
Choose the letter with the smaller LCM to eliminate; it keeps the numbers small.
Always find the second unknown by substituting into the SIMPLER original equation, not into one you have multiplied.
Check in BOTH original equations — and for a question involving fractions or brackets, check in the originals rather than the cleared versions, so that the clearing step is tested too.
Verifying means both equations. A pair satisfying only one is not a solution.
To find a constant, substitute the given solution — with one constant you get one equation, and with two constants you get a pair to solve.
And state the answer as a pair, and , rather than leaving two isolated numbers at the foot of the page.
Did you know
Why two lines meeting at a point is the whole story
Every pair of simultaneous linear equations is two straight lines, and solving the pair is finding where they cross. That one picture explains everything the algebra can do — including the cases where it fails.
Two lines in a plane can be arranged in exactly three ways.
They can cross once. This is the ordinary case, there is a single solution, and both methods find it.
They can be parallel and distinct. Then they never meet, and there is no solution. In the algebra this shows up as an absurdity: try to solve and by subtraction and you get , which is false. The method has not broken; it is telling you that no such point exists.
Or they can be the same line drawn twice. Then every point on it satisfies both equations, and there are infinitely many solutions. Try to solve and and you get — true, but empty of information, because the second equation is only a disguised copy of the other one and supplies nothing new.
So when a calculation collapses to or to something plainly false, that is a result rather than a mistake, and it is worth recognising rather than starting again.
You can tell which case you are in before solving anything, just by looking at the coefficients of and .
If , the lines have different gradients and cross once. If all three ratios , and are equal, the equations are multiples of each other and the lines coincide. And if the first two ratios are equal but the third differs, the lines are parallel and there is no solution.
Which means the real content of "two equations pin down two unknowns" has a condition attached: they must be two genuinely different equations, not one equation written twice. And the next part of this chapter is where that condition is put to work.
Two lines in a plane can be arranged in exactly three ways.
They can cross once. This is the ordinary case, there is a single solution, and both methods find it.
They can be parallel and distinct. Then they never meet, and there is no solution. In the algebra this shows up as an absurdity: try to solve and by subtraction and you get , which is false. The method has not broken; it is telling you that no such point exists.
Or they can be the same line drawn twice. Then every point on it satisfies both equations, and there are infinitely many solutions. Try to solve and and you get — true, but empty of information, because the second equation is only a disguised copy of the other one and supplies nothing new.
So when a calculation collapses to or to something plainly false, that is a result rather than a mistake, and it is worth recognising rather than starting again.
You can tell which case you are in before solving anything, just by looking at the coefficients of and .
If , the lines have different gradients and cross once. If all three ratios , and are equal, the equations are multiples of each other and the lines coincide. And if the first two ratios are equal but the third differs, the lines are parallel and there is no solution.
Which means the real content of "two equations pin down two unknowns" has a condition attached: they must be two genuinely different equations, not one equation written twice. And the next part of this chapter is where that condition is put to work.
Exam relevance
Why does JEE Main still use elimination in three variables?
Because the method scales. Two equations in two unknowns is the smallest case of a system, and the same elimination becomes the standard technique for three.
This is the foundation for Class 11 Mathematics Straight Lines and Class 12 Matrices and Determinants, examined in JEE Main. Straight Lines makes the picture explicit — a linear equation is a line, and solving a pair is finding the point of intersection. Questions asking for the intersection of two given lines, or for the line through that intersection, are standard, and they are this chapter with geometric wording.
The three-case analysis becomes the consistency condition. Class 12 Determinants gives Cramer's rule and states the cases formally: a unique solution when the determinant is non-zero, and no solution or infinitely many when it is zero, distinguished by further determinants. The coefficient-ratio test of the fun fact is exactly that condition written out for two variables, and questions on the value of a constant for which a system has no solution or infinitely many are a recurring JEE Main type.
Elimination becomes row reduction. Class 12 Matrices writes a system as and solves it by the inverse matrix method or by row operations — and a row operation is precisely "multiply an equation and add it to another", which is what elimination does. So the technique is not replaced later; it is formalised and extended to three variables.
Word problems built on these methods run right through the syllabus. Class 10 and 11 set problems on ages, two-digit numbers, boats and streams, fractions, and work rates — all reduced to a pair of linear equations. The framing changes and the solving does not, which is why speed and accuracy here pay off later.
Substitution reappears everywhere. Solving a linear equation together with a quadratic — a line meeting a circle or a parabola — is done by substituting the linear equation into the other, exactly as on this page. Questions on the intersection of a line and a conic are standard in JEE Main, and the number of solutions then tells you whether the line cuts, touches or misses the curve.
Linear programming depends on the same skill. Class 12 Linear Programming finds the corner points of a feasible region, and every corner is the intersection of two boundary lines — found by solving a pair of simultaneous equations. A graph with five constraints needs this done several times, quickly.
What the questions look like. For board work, expect solve a given pair by substitution, solve a given pair by elimination, solve a pair after clearing fractions or brackets, verify a given solution, and find an unknown constant given the solution. Equations should be numbered and the check shown. For JEE Main, expect intersection points, consistency conditions with a parameter, matrix and determinant methods, and line-meets-curve substitutions.
How board and competitive emphasis differ. A board paper rewards the method set out line by line with both checks. A competitive paper assumes the solving and asks for the value of a constant that makes a system inconsistent, or for a point of intersection used in a further calculation.
The single trap that costs the most marks. Checking the answer in only one of the two equations. A pair of values satisfying one equation lies somewhere on one of the two lines — which is true of infinitely many pairs and proves nothing. Only a pair satisfying both is the intersection point. The defence is to write the two checks as two separate lines with a tick against each, because an arithmetic slip made while eliminating will very often produce values that still satisfy the equation you substituted back into, and the other equation is the one that exposes it.
This is the foundation for Class 11 Mathematics Straight Lines and Class 12 Matrices and Determinants, examined in JEE Main. Straight Lines makes the picture explicit — a linear equation is a line, and solving a pair is finding the point of intersection. Questions asking for the intersection of two given lines, or for the line through that intersection, are standard, and they are this chapter with geometric wording.
The three-case analysis becomes the consistency condition. Class 12 Determinants gives Cramer's rule and states the cases formally: a unique solution when the determinant is non-zero, and no solution or infinitely many when it is zero, distinguished by further determinants. The coefficient-ratio test of the fun fact is exactly that condition written out for two variables, and questions on the value of a constant for which a system has no solution or infinitely many are a recurring JEE Main type.
Elimination becomes row reduction. Class 12 Matrices writes a system as and solves it by the inverse matrix method or by row operations — and a row operation is precisely "multiply an equation and add it to another", which is what elimination does. So the technique is not replaced later; it is formalised and extended to three variables.
Word problems built on these methods run right through the syllabus. Class 10 and 11 set problems on ages, two-digit numbers, boats and streams, fractions, and work rates — all reduced to a pair of linear equations. The framing changes and the solving does not, which is why speed and accuracy here pay off later.
Substitution reappears everywhere. Solving a linear equation together with a quadratic — a line meeting a circle or a parabola — is done by substituting the linear equation into the other, exactly as on this page. Questions on the intersection of a line and a conic are standard in JEE Main, and the number of solutions then tells you whether the line cuts, touches or misses the curve.
Linear programming depends on the same skill. Class 12 Linear Programming finds the corner points of a feasible region, and every corner is the intersection of two boundary lines — found by solving a pair of simultaneous equations. A graph with five constraints needs this done several times, quickly.
What the questions look like. For board work, expect solve a given pair by substitution, solve a given pair by elimination, solve a pair after clearing fractions or brackets, verify a given solution, and find an unknown constant given the solution. Equations should be numbered and the check shown. For JEE Main, expect intersection points, consistency conditions with a parameter, matrix and determinant methods, and line-meets-curve substitutions.
How board and competitive emphasis differ. A board paper rewards the method set out line by line with both checks. A competitive paper assumes the solving and asks for the value of a constant that makes a system inconsistent, or for a point of intersection used in a further calculation.
The single trap that costs the most marks. Checking the answer in only one of the two equations. A pair of values satisfying one equation lies somewhere on one of the two lines — which is true of infinitely many pairs and proves nothing. Only a pair satisfying both is the intersection point. The defence is to write the two checks as two separate lines with a tick against each, because an arithmetic slip made while eliminating will very often produce values that still satisfy the equation you substituted back into, and the other equation is the one that exposes it.
Key takeaways
Simultaneous linear equations: quick revision
- One equation in two unknowns has infinitely many solutions, forming a straight line. Two equations pin down the single point where the lines cross.
- Simultaneous means both must hold at once — a pair satisfying only one is not a solution.
- Substitution: make one letter the subject of the simpler equation, substitute into the other, solve for one unknown, then back-substitute.
- Worked: and . From the second, ; then gives , and .
- **Choose a letter whose coefficient is or to be the subject — otherwise fractions follow you through every line.
- Put the substituted expression in brackets**: , never .
- Elimination: multiply so that one letter's coefficients match in size, then subtract if the signs are the same and add if they are opposite.
- Worked: and . Multiply the second by to get , then add: , , and .
- Worked with both multiplied: and . Multiply by and to get and , then add: , , .
- Multiply every term including the right-hand side, and eliminate the letter with the smaller LCM.
- Clear fractions by multiplying through by the LCM of the denominators. and become and , giving , .
- Expand brackets and collect before solving. becomes , and becomes , giving , .
- A minus outside a bracket changes every sign inside: .
- Check in the ORIGINAL equations, not the cleared ones, so that the clearing step is tested too.
- Verifying requires BOTH equations. satisfies and fails , so it is not a solution.
- To find one constant, substitute the given solution: in gives , so .
- To find two constants, substitute into both equations and solve the resulting pair: in and gives and , so , .
- The unknowns can be any letters — in that last problem and were the known numbers.
- Three geometric cases: lines crossing once (one solution), parallel (no solution, algebra gives something false), or coincident (infinitely many, algebra gives ).
- Number your equations and refer to the numbers — the method marks depend on it.
Solve any pair by both methods and confirm you get the same point — then check it in both originals, because that is the only check that proves anything.
- Simultaneous means both must hold at once — a pair satisfying only one is not a solution.
- Substitution: make one letter the subject of the simpler equation, substitute into the other, solve for one unknown, then back-substitute.
- Worked: and . From the second, ; then gives , and .
- **Choose a letter whose coefficient is or to be the subject — otherwise fractions follow you through every line.
- Put the substituted expression in brackets**: , never .
- Elimination: multiply so that one letter's coefficients match in size, then subtract if the signs are the same and add if they are opposite.
- Worked: and . Multiply the second by to get , then add: , , and .
- Worked with both multiplied: and . Multiply by and to get and , then add: , , .
- Multiply every term including the right-hand side, and eliminate the letter with the smaller LCM.
- Clear fractions by multiplying through by the LCM of the denominators. and become and , giving , .
- Expand brackets and collect before solving. becomes , and becomes , giving , .
- A minus outside a bracket changes every sign inside: .
- Check in the ORIGINAL equations, not the cleared ones, so that the clearing step is tested too.
- Verifying requires BOTH equations. satisfies and fails , so it is not a solution.
- To find one constant, substitute the given solution: in gives , so .
- To find two constants, substitute into both equations and solve the resulting pair: in and gives and , so , .
- The unknowns can be any letters — in that last problem and were the known numbers.
- Three geometric cases: lines crossing once (one solution), parallel (no solution, algebra gives something false), or coincident (infinitely many, algebra gives ).
- Number your equations and refer to the numbers — the method marks depend on it.
Solve any pair by both methods and confirm you get the same point — then check it in both originals, because that is the only check that proves anything.