Three Ratios Tell You How Many Solutions a Pair of Equations Has
Compare two lines using the ratios of their coefficients, decide whether a pair of equations has one solution, none, or infinitely many, solve graphically, find an unknown k from a consistency condition, and get the area of a triangle formed by two lines and an axis.
How can you tell a pair of equations has no solution without solving it?
Two straight lines in a plane can do only three things. They cross at one point, they lie exactly on top of each other, or they never meet at all. There is no fourth possibility, and each case corresponds to a different answer for the pair of equations.
- Lines that cross once — exactly one solution, the point of intersection
- Lines that coincide — infinitely many solutions, since every point on the line satisfies both
- Lines that are parallel and distinct — no solution at all
The striking part is that you can tell which case you are in by looking only at the coefficients. Write both equations in the form , form three ratios, and compare them. No graph, no substitution, no elimination — and questions worth three marks are answered in two lines.
That test then does more work than classification. A great many questions supply a letter in place of one coefficient and ask for the value that makes the pair have no solution, or infinitely many. Those are one-line equations in the unknown letter, once you know which ratios must be equal.
The chapter also keeps the graph, and for good reason. Drawing the two lines gives you the solution directly, and it lets you answer a kind of question algebra cannot: what is the area of the triangle formed by two lines and an axis. That needs the vertices, and the vertices come from the graph.
This page covers the first part of the CBSE Class 10 Maths chapter on a pair of linear equations in two variables: graphical solution, consistency, the ratio test, and finding an unknown coefficient from a consistency condition.
- Lines that cross once — exactly one solution, the point of intersection
- Lines that coincide — infinitely many solutions, since every point on the line satisfies both
- Lines that are parallel and distinct — no solution at all
The striking part is that you can tell which case you are in by looking only at the coefficients. Write both equations in the form , form three ratios, and compare them. No graph, no substitution, no elimination — and questions worth three marks are answered in two lines.
That test then does more work than classification. A great many questions supply a letter in place of one coefficient and ask for the value that makes the pair have no solution, or infinitely many. Those are one-line equations in the unknown letter, once you know which ratios must be equal.
The chapter also keeps the graph, and for good reason. Drawing the two lines gives you the solution directly, and it lets you answer a kind of question algebra cannot: what is the area of the triangle formed by two lines and an axis. That needs the vertices, and the vertices come from the graph.
This page covers the first part of the CBSE Class 10 Maths chapter on a pair of linear equations in two variables: graphical solution, consistency, the ratio test, and finding an unknown coefficient from a consistency condition.
Formula
What does the ratio test say, and how do you apply it?
**Write both equations in the standard form and compare the ratios of the coefficients of , of , and of the constants.**
For the pair
the three cases are:
The vocabulary matters as much as the arithmetic. A pair with at least one solution is consistent; a pair with no solution is inconsistent. A coincident pair is sometimes called dependent, because one equation is just a multiple of the other.
Worked example 1 — a unique solution. Examine the pair and , and solve it if possible.
In standard form, , , and , , .
These are unequal, so the lines intersect and there is exactly one solution.
Solving: from the first equation . Substituting into the second,
and then . **The solution is , .
Check in both equations**: , and . Both satisfied.
Worked example 2 — infinitely many solutions. Examine and .
In standard form the constants are and , so
All three ratios are equal, so the lines coincide and there are infinitely many solutions. Indeed the second equation is simply twice the first, which is why it adds no new information.
Worked example 3 — no solution. Examine and .
The first two ratios are equal but the third is different, so the lines are parallel and the pair is inconsistent — no solution.
Notice how close examples 2 and 3 look. Both have proportional and coefficients; only the constant tells them apart. That is why the third ratio must always be computed and never assumed, and it is the difference between "infinitely many" and "none" — two answers that could not be further apart.
One point of care with the constants. Move everything to the left-hand side before reading off and . Writing the pair as and comparing with happens to give the same ratio here, but with mixed signs it will not, and the standard form removes the risk entirely.
For the pair
the three cases are:
The vocabulary matters as much as the arithmetic. A pair with at least one solution is consistent; a pair with no solution is inconsistent. A coincident pair is sometimes called dependent, because one equation is just a multiple of the other.
Worked example 1 — a unique solution. Examine the pair and , and solve it if possible.
In standard form, , , and , , .
These are unequal, so the lines intersect and there is exactly one solution.
Solving: from the first equation . Substituting into the second,
and then . **The solution is , .
Check in both equations**: , and . Both satisfied.
Worked example 2 — infinitely many solutions. Examine and .
In standard form the constants are and , so
All three ratios are equal, so the lines coincide and there are infinitely many solutions. Indeed the second equation is simply twice the first, which is why it adds no new information.
Worked example 3 — no solution. Examine and .
The first two ratios are equal but the third is different, so the lines are parallel and the pair is inconsistent — no solution.
Notice how close examples 2 and 3 look. Both have proportional and coefficients; only the constant tells them apart. That is why the third ratio must always be computed and never assumed, and it is the difference between "infinitely many" and "none" — two answers that could not be further apart.
One point of care with the constants. Move everything to the left-hand side before reading off and . Writing the pair as and comparing with happens to give the same ratio here, but with mixed signs it will not, and the standard form removes the risk entirely.
How do you find the value of k for a given number of solutions?
**Decide which ratio condition the question describes, write it as an equation in , and solve. Then check the remaining ratio, because sometimes it eliminates one of your answers.
Worked example 1 — infinitely many solutions.** Find so that the pair and has infinitely many solutions.
All three ratios must be equal:
From the first two:
Now use the third ratio to decide between them. Taking the first and third,
**So only**, and is rejected.
**Check with .** The equations become and , and the second is exactly twice the first. Coincident, as required. With the equations would be and , where the coefficient ratios are but the constant ratio is — parallel, not coincident, so the rejection was correct.
That double-check is the whole lesson of this example. Two of the three ratios give a quadratic in , and the third ratio is what selects the right root.
Worked example 2 — infinitely many, with the answer unique from the start. Find so that and has infinitely many solutions.
In standard form: and . Equating the first two ratios,
**Check the third ratio with :**
**All three equal, so is correct.
Worked example 3 — no solution.** Find so that and has no solution.
Now the condition is that the first two ratios are equal while the third is different.
Now confirm the third ratio really is different. With the equations are and , so
The first two are equal and the third is not, so the lines are parallel and there is no solution. Hence .
Notice that the confirmation was essential here too. Had the third ratio also come out as , the lines would have coincided and would have been the wrong answer for this question. The inequality is part of the condition, not an afterthought.
Worked example 4 — a unique solution. Find the values of for which and has a unique solution.
A unique solution needs the first two ratios to be unequal:
**So every value of except gives a unique solution.** Questions of this type have an inequality for an answer, not a number, and writing a single value would be wrong.
Worked example 1 — infinitely many solutions.** Find so that the pair and has infinitely many solutions.
All three ratios must be equal:
From the first two:
Now use the third ratio to decide between them. Taking the first and third,
**So only**, and is rejected.
**Check with .** The equations become and , and the second is exactly twice the first. Coincident, as required. With the equations would be and , where the coefficient ratios are but the constant ratio is — parallel, not coincident, so the rejection was correct.
That double-check is the whole lesson of this example. Two of the three ratios give a quadratic in , and the third ratio is what selects the right root.
Worked example 2 — infinitely many, with the answer unique from the start. Find so that and has infinitely many solutions.
In standard form: and . Equating the first two ratios,
**Check the third ratio with :**
**All three equal, so is correct.
Worked example 3 — no solution.** Find so that and has no solution.
Now the condition is that the first two ratios are equal while the third is different.
Now confirm the third ratio really is different. With the equations are and , so
The first two are equal and the third is not, so the lines are parallel and there is no solution. Hence .
Notice that the confirmation was essential here too. Had the third ratio also come out as , the lines would have coincided and would have been the wrong answer for this question. The inequality is part of the condition, not an afterthought.
Worked example 4 — a unique solution. Find the values of for which and has a unique solution.
A unique solution needs the first two ratios to be unequal:
**So every value of except gives a unique solution.** Questions of this type have an inequality for an answer, not a number, and writing a single value would be wrong.
How do you solve a pair graphically and find the triangle's area?
Plot each line from two or three points, read the intersection, and then use the intercepts as the other vertices of the triangle.
The method, step by step.
- For each equation, make a small table of values. The easiest points are the intercepts: put to get the -intercept and to get the -intercept
- Plot at least three points per line so a plotting slip shows up as a point off the line
- Draw both lines and read the coordinates of the point where they cross
- Verify the intersection by substituting into both original equations
Worked example — the full triangle question. Draw the graphs of and , and find the area of the triangle formed by these lines and the -axis.
**Step 1 — where each line meets the -axis.** On the -axis, .
For the first line: , so . **The vertex is .**
For the second line: , so . **The vertex is .
Step 2 — where the two lines meet each other.** From the first equation, . Substituting into the second,
and then . **The third vertex is .
Check in both equations**: , and . Both satisfied.
Step 3 — the area. The base lies along the -axis between and :
The height is the perpendicular distance from to the -axis, which is simply its -coordinate:
Two things made that easy, and both are worth noticing. The base lay along an axis, so its length was just the difference of the -coordinates. And the height was the -coordinate of the opposite vertex, **because the distance from a point to the -axis is its -coordinate.** Whenever a triangle has a side on an axis, the area needs no formula beyond .
**If the triangle is formed with the -axis instead, everything swaps: the base is the difference of the -intercepts, and the height is the -coordinate of the intersection point.
What the graph shows in the other two cases.
- An inconsistent pair gives two parallel lines that never meet, and the graph itself is the answer — there is no intersection to read
- A dependent pair gives one line drawn twice, so every point on it is a solution. If your two tables of values produce the same line, check whether one equation is a multiple of the other** before assuming you made a plotting error
The method, step by step.
- For each equation, make a small table of values. The easiest points are the intercepts: put to get the -intercept and to get the -intercept
- Plot at least three points per line so a plotting slip shows up as a point off the line
- Draw both lines and read the coordinates of the point where they cross
- Verify the intersection by substituting into both original equations
Worked example — the full triangle question. Draw the graphs of and , and find the area of the triangle formed by these lines and the -axis.
**Step 1 — where each line meets the -axis.** On the -axis, .
For the first line: , so . **The vertex is .**
For the second line: , so . **The vertex is .
Step 2 — where the two lines meet each other.** From the first equation, . Substituting into the second,
and then . **The third vertex is .
Check in both equations**: , and . Both satisfied.
Step 3 — the area. The base lies along the -axis between and :
The height is the perpendicular distance from to the -axis, which is simply its -coordinate:
Two things made that easy, and both are worth noticing. The base lay along an axis, so its length was just the difference of the -coordinates. And the height was the -coordinate of the opposite vertex, **because the distance from a point to the -axis is its -coordinate.** Whenever a triangle has a side on an axis, the area needs no formula beyond .
**If the triangle is formed with the -axis instead, everything swaps: the base is the difference of the -intercepts, and the height is the -coordinate of the intersection point.
What the graph shows in the other two cases.
- An inconsistent pair gives two parallel lines that never meet, and the graph itself is the answer — there is no intersection to read
- A dependent pair gives one line drawn twice, so every point on it is a solution. If your two tables of values produce the same line, check whether one equation is a multiple of the other** before assuming you made a plotting error
Exam tip
What are the marks in a consistency question actually for?
Standard form first, all three ratios second, and the conclusion stated in words. Skipping any of the three costs marks even when the final answer is right.
- **Rewrite both equations as before reading off any coefficient, with the constant on the left and its correct sign
- Compute all three ratios**, even when the first two already differ. Showing costs one line and proves you checked
- State the conclusion in words: "the lines intersect at one point, so the pair is consistent with a unique solution"
- For infinitely many solutions, all three ratios must be equal — and when two ratios give a quadratic in , use the third to reject the wrong root
- For no solution, the first two must be equal and the third different — confirm the inequality, do not assume it
- For a unique solution, the answer is an inequality such as , not a value
- On a graph, plot three points per line and label both lines with their equations
- For an area question, list the three vertices before computing anything
The misconception to name. "No solution" and "infinitely many solutions" are not near-misses of each other — they are opposite ends of the scale, and only the constant ratio distinguishes them. The pair with has none while with has infinitely many, and the coefficients look equally proportional in both cases.
A second trap. Reading the constant ratio with the wrong sign. Writing and comparing against works only because both are positive; once one constant is negative the unmoved form gives the wrong ratio. Move everything to the left first, every time.
- **Rewrite both equations as before reading off any coefficient, with the constant on the left and its correct sign
- Compute all three ratios**, even when the first two already differ. Showing costs one line and proves you checked
- State the conclusion in words: "the lines intersect at one point, so the pair is consistent with a unique solution"
- For infinitely many solutions, all three ratios must be equal — and when two ratios give a quadratic in , use the third to reject the wrong root
- For no solution, the first two must be equal and the third different — confirm the inequality, do not assume it
- For a unique solution, the answer is an inequality such as , not a value
- On a graph, plot three points per line and label both lines with their equations
- For an area question, list the three vertices before computing anything
The misconception to name. "No solution" and "infinitely many solutions" are not near-misses of each other — they are opposite ends of the scale, and only the constant ratio distinguishes them. The pair with has none while with has infinitely many, and the coefficients look equally proportional in both cases.
A second trap. Reading the constant ratio with the wrong sign. Writing and comparing against works only because both are positive; once one constant is negative the unmoved form gives the wrong ratio. Move everything to the left first, every time.
Did you know
Why does a pair of equations sometimes carry no new information?
Look at and side by side. The second is the first multiplied by two. So it is not a second piece of information at all — it is the same statement said louder.
That is why the pair has infinitely many solutions. You were given one equation twice and asked to find two unknowns, and one equation can never pin down two quantities. Any point on the line satisfies both, and there are infinitely many such points.
The everyday version of this is worth recognising, because word problems are built on it. Suppose a problem tells you that two pens and three notebooks cost ₹, and then tells you that four pens and six notebooks cost ₹. The second sentence is just the first doubled, so you still cannot find the price of a pen. The problem has no unique answer, and the ratio test says so in one line.
Now change one digit and everything reverses. If four pens and six notebooks cost ₹ instead, the two statements contradict each other: doubling the first order must double its cost, so ₹ is the only consistent figure. There is now no solution at all — the situation described is impossible. A parallel pair of lines is an impossible story, and a coincident pair is a repeated one.
Which makes the three cases very intuitive once named.
- Two genuinely different pieces of information — one answer
- The same information twice — infinitely many answers
- Two pieces of information that contradict — no answer
And it explains why the third ratio is the decisive one. The coefficient ratios tell you whether the two statements are about the same combination of and . If they are, the constants decide whether the statements agree or disagree. The first two ratios ask "are you saying the same kind of thing?" and the third asks "and do you agree?"
One last thought about how often this matters outside the chapter. Any time a real problem produces more equations than unknowns, the same question arises: are the extra equations new information, repetition, or contradiction? Engineers and statisticians deal with exactly this, and the language they use — consistent, inconsistent, dependent — is the language you are learning here on two lines in a plane.
That is why the pair has infinitely many solutions. You were given one equation twice and asked to find two unknowns, and one equation can never pin down two quantities. Any point on the line satisfies both, and there are infinitely many such points.
The everyday version of this is worth recognising, because word problems are built on it. Suppose a problem tells you that two pens and three notebooks cost ₹, and then tells you that four pens and six notebooks cost ₹. The second sentence is just the first doubled, so you still cannot find the price of a pen. The problem has no unique answer, and the ratio test says so in one line.
Now change one digit and everything reverses. If four pens and six notebooks cost ₹ instead, the two statements contradict each other: doubling the first order must double its cost, so ₹ is the only consistent figure. There is now no solution at all — the situation described is impossible. A parallel pair of lines is an impossible story, and a coincident pair is a repeated one.
Which makes the three cases very intuitive once named.
- Two genuinely different pieces of information — one answer
- The same information twice — infinitely many answers
- Two pieces of information that contradict — no answer
And it explains why the third ratio is the decisive one. The coefficient ratios tell you whether the two statements are about the same combination of and . If they are, the constants decide whether the statements agree or disagree. The first two ratios ask "are you saying the same kind of thing?" and the third asks "and do you agree?"
One last thought about how often this matters outside the chapter. Any time a real problem produces more equations than unknowns, the same question arises: are the extra equations new information, repetition, or contradiction? Engineers and statisticians deal with exactly this, and the language they use — consistent, inconsistent, dependent — is the language you are learning here on two lines in a plane.
Exam relevance
How does the ratio test prepare you for JEE?
This is foundation work for Class 11 Straight Lines and Class 12 Matrices and Determinants, examined in JEE Main and JEE Advanced.
Where the three cases lead. Class 12 replaces the ratio test with the determinant of the coefficient matrix. A pair has a unique solution exactly when , which is precisely your condition cleared of fractions. The determinant form is strictly better, because it does not break down when a coefficient is zero and the ratio becomes undefined — and recognising that it is the same test is what makes the Class 12 chapter easy rather than new.
Where the consistency vocabulary leads. Class 12 extends consistent, inconsistent and dependent to systems of three equations in three unknowns, solved by Cramer's rule or by the matrix inverse. The three-way classification is identical, and the geometry becomes three planes meeting in a point, a line, or not at all.
Where the graphical picture leads. Class 11 Straight Lines gives the conditions for two lines to be parallel or perpendicular in terms of their slopes, and the condition for three lines to be concurrent as a determinant. The parallel case you meet here is the slope condition in disguise, since equal coefficient ratios mean equal slopes.
**Where the find- questions lead. JEE Main sets exactly this family, with the system written as a determinant set to zero. The habit of solving two ratios and then testing the third against the remaining condition survives unchanged — in matrix language it becomes checking the rank of the coefficient matrix against the rank of the augmented matrix.
Where the area calculation leads. Class 11 gives the general formula for the area of a triangle from three vertices, and Class 12 writes it as a determinant. The shortcut you use here works because a side lies on an axis, and it stays the fastest route whenever that happens.
Question types to expect.** At this level: classify a pair, solve graphically, find for a stated number of solutions, and find the area of a triangle formed with an axis. In competitive papers: determinant conditions, three-equation systems, concurrency of lines, and problems where the number of solutions must be stated for all values of a parameter.
The single trap that costs marks. Stopping after the first two ratios when the question asks about infinitely many or no solutions. Two ratios cannot distinguish those two cases, and since they are opposite answers, the third ratio is the entire content of the question.
A second trap. Accepting both roots of the quadratic that comes out of a find- question. **The example shows why**: satisfies two ratios but makes the lines parallel rather than coincident, so it answers a different question. In Class 12 the same rejection is done by comparing ranks, and the reasoning is the same.
Board versus competitive emphasis. The CBSE paper marks the standard form, the three ratios, the stated conclusion and the graph; a competitive paper marks the determinant condition and the parameter range. The transferable habit is asking whether the second equation adds information, repeats it, or contradicts it — because that question is what rank and determinant are measuring, and it is already answerable with the three ratios you have here.
Where the three cases lead. Class 12 replaces the ratio test with the determinant of the coefficient matrix. A pair has a unique solution exactly when , which is precisely your condition cleared of fractions. The determinant form is strictly better, because it does not break down when a coefficient is zero and the ratio becomes undefined — and recognising that it is the same test is what makes the Class 12 chapter easy rather than new.
Where the consistency vocabulary leads. Class 12 extends consistent, inconsistent and dependent to systems of three equations in three unknowns, solved by Cramer's rule or by the matrix inverse. The three-way classification is identical, and the geometry becomes three planes meeting in a point, a line, or not at all.
Where the graphical picture leads. Class 11 Straight Lines gives the conditions for two lines to be parallel or perpendicular in terms of their slopes, and the condition for three lines to be concurrent as a determinant. The parallel case you meet here is the slope condition in disguise, since equal coefficient ratios mean equal slopes.
**Where the find- questions lead. JEE Main sets exactly this family, with the system written as a determinant set to zero. The habit of solving two ratios and then testing the third against the remaining condition survives unchanged — in matrix language it becomes checking the rank of the coefficient matrix against the rank of the augmented matrix.
Where the area calculation leads. Class 11 gives the general formula for the area of a triangle from three vertices, and Class 12 writes it as a determinant. The shortcut you use here works because a side lies on an axis, and it stays the fastest route whenever that happens.
Question types to expect.** At this level: classify a pair, solve graphically, find for a stated number of solutions, and find the area of a triangle formed with an axis. In competitive papers: determinant conditions, three-equation systems, concurrency of lines, and problems where the number of solutions must be stated for all values of a parameter.
The single trap that costs marks. Stopping after the first two ratios when the question asks about infinitely many or no solutions. Two ratios cannot distinguish those two cases, and since they are opposite answers, the third ratio is the entire content of the question.
A second trap. Accepting both roots of the quadratic that comes out of a find- question. **The example shows why**: satisfies two ratios but makes the lines parallel rather than coincident, so it answers a different question. In Class 12 the same rejection is done by comparing ranks, and the reasoning is the same.
Board versus competitive emphasis. The CBSE paper marks the standard form, the three ratios, the stated conclusion and the graph; a competitive paper marks the determinant condition and the parameter range. The transferable habit is asking whether the second equation adds information, repeats it, or contradicts it — because that question is what rank and determinant are measuring, and it is already answerable with the three ratios you have here.
Key takeaways
What should you carry into the algebraic methods?
Three cases, three ratios and one area shortcut.
- Two lines either intersect once, coincide, or are parallel — there is no fourth possibility
- **Write both equations as before reading off any coefficient
- — intersecting lines, one solution, consistent
- — coincident lines, infinitely many solutions, consistent and dependent
- — parallel lines, no solution, inconsistent
- Only the third ratio separates "infinitely many" from "none", so always compute it
- with ** gives the unique solution
- ** with is dependent; with is inconsistent
- For infinitely many solutions, two ratios give a quadratic in and the third selects the root** — with gives , not
- ** with ** is dependent when
- ** with ** has no solution when
- A unique-solution question has an inequality for an answer, such as
- **For a triangle with a side on the -axis**, the base is the difference of the -intercepts and the height is the -coordinate of the third vertex — so , and give an area of square units
The quickest way to know this is secure is to invent the trap yourself. Write down any equation, double it, then change only the constant — and predict, before computing a single ratio, whether the pair you have just built has infinitely many solutions or none.
- Two lines either intersect once, coincide, or are parallel — there is no fourth possibility
- **Write both equations as before reading off any coefficient
- — intersecting lines, one solution, consistent
- — coincident lines, infinitely many solutions, consistent and dependent
- — parallel lines, no solution, inconsistent
- Only the third ratio separates "infinitely many" from "none", so always compute it
- with ** gives the unique solution
- ** with is dependent; with is inconsistent
- For infinitely many solutions, two ratios give a quadratic in and the third selects the root** — with gives , not
- ** with ** is dependent when
- ** with ** has no solution when
- A unique-solution question has an inequality for an answer, such as
- **For a triangle with a side on the -axis**, the base is the difference of the -intercepts and the height is the -coordinate of the third vertex — so , and give an area of square units
The quickest way to know this is secure is to invent the trap yourself. Write down any equation, double it, then change only the constant — and predict, before computing a single ratio, whether the pair you have just built has infinitely many solutions or none.