NCERT Exemplar Solutions - Class 10 - Mathematics - Unit 3: Pair of Linear Equations in Two Variables

Exercise 3.2

How to decide: First write each equation in the form \(a_1x + b_1y + c_1 = 0\) and \(a_2x + b_2y + c_2 = 0\).

If \(\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2}\) but \(\dfrac{c_1}{c_2}\) is different, the lines are parallel.

Parallel lines never meet, so there is no solution.

(i) Convert both to standard form.

\(2x + 4y - 3 = 0\)

\(6x + 12y - 6 = 0\)

Compute ratios one by one.

\(\dfrac{a_1}{a_2} = \dfrac{2}{6} = \dfrac{1}{3}\)

\(\dfrac{b_1}{b_2} = \dfrac{4}{12} = \dfrac{1}{3}\)

\(\dfrac{c_1}{c_2} = \dfrac{-3}{-6} = \dfrac{1}{2}\)

The first two ratios are equal, but the third is different.

Therefore, the lines are parallel ⇒ no solution.

(ii) Write in standard form.

\(x - 2y = 0\)

\(-2x + y = 0\)

Here \(\dfrac{a_1}{a_2} = \dfrac{1}{-2}\) and \(\dfrac{b_1}{b_2} = \dfrac{-2}{1}\).

These are not equal, so the lines intersect at one point (unique solution).

In fact, using the original form: from \(x = 2y\) and \(y = 2x\), substitute to get \(x = 2(2x)\).

That is \(x = 4x\) ⇒ \(3x = 0\) ⇒ \(x = 0\), and hence \(y = 0\).

So there is a unique solution, not “no solution”.

(iii) Standard form first.

\(3x + y - 3 = 0\)

\(\dfrac{2}{3}x + \dfrac{1}{2}y - 2 = 0\)

Compare ratios.

\(\dfrac{a_1}{a_2} = \dfrac{3}{\,2/3\,} = \dfrac{9}{2}\)

\(\dfrac{b_1}{b_2} = \dfrac{1}{\,1/2\,} = 2\)

Since these are unequal, the lines meet at one point (unique solution).

Therefore, the answer is “No” to “no solution”.

Concept: Two equations represent coincident lines if and only if

\(\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2}\).

Here, each equation is written as \(a_1x + b_1y + c_1 = 0\) and \(a_2x + b_2y + c_2 = 0\).

(i)

First equation: \(3x + \dfrac{1}{7}y - 3 = 0\)

Second equation: \(7x + 3y - 7 = 0\)

Now, compare the ratios:

\(\dfrac{a_1}{a_2} = \dfrac{3}{7}\)

\(\dfrac{b_1}{b_2} = \dfrac{1/7}{3} = \dfrac{1}{21}\)

Since \(\dfrac{3}{7} \ne \dfrac{1}{21}\), the lines are not coincident.

(ii)

First equation: \(-2x - 3y - 1 = 0\)

Second equation: \(4x + 6y + 2 = 0\)

If we multiply the first equation by \(-2\), we get:

\(4x + 6y + 2 = 0\)

which is exactly the second equation.

Thus, all three ratios are equal, so the lines are coincident.

(iii)

First equation: \(\dfrac{x}{2} + \dfrac{y}{5} + \dfrac{5}{16} = 0\)

Multiply throughout by 80 to clear fractions:

\(40x + 16y + 25 = 0\)

Second equation: \(4x + 8y + \dfrac{5}{4} = 0\)

Multiply throughout by 4:

\(16x + 32y + 5 = 0\)

Now compare ratios:

\(\dfrac{a_1}{a_2} = \dfrac{40}{16} = 2.5\)

\(\dfrac{b_1}{b_2} = \dfrac{16}{32} = 0.5\)

Since these are not equal, the lines are not coincident.

Final Answer:

(i) No, (ii) Yes, (iii) No

Rule to check: If

\(\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} \ne \dfrac{c_1}{c_2}\)

the lines are parallel ⇒ inconsistent (no solution).

Otherwise, they are consistent (they meet or coincide).

(i) Put both in standard form.

\(-3x - 4y - 12 = 0\)

\(3x + 4y - 12 = 0\)

Compute the three ratios.

\(\dfrac{a_1}{a_2} = \dfrac{-3}{3} = -1\)

\(\dfrac{b_1}{b_2} = \dfrac{-4}{4} = -1\)

\(\dfrac{c_1}{c_2} = \dfrac{-12}{-12} = 1\)

Here the first two ratios are equal, but the third is different.

So the lines are parallel ⇒ inconsistent.

(ii) Clear fractions.

Multiply both equations by 5.

\(3x - 5y = 60\)

\(x - 15y = 80\)

Compare ratios.

\(\dfrac{a_1}{a_2} = \dfrac{3}{1}\)

\(\dfrac{b_1}{b_2} = \dfrac{-5}{-15} = \dfrac{1}{3}\)

These are not equal, so the lines intersect ⇒ consistent.

(iii) Arrange both equations.

\(2ax + by - a = 0\)

\(4ax + 2by - 2a = 0\)

Notice that the second is exactly twice the first.

Multiply the first by 2 to get the second.

Hence all three ratios are equal, so the lines are coincident.

Therefore, they are consistent with infinitely many solutions.

(iv) Expand the second equation.

\(2(2x + 6y) = 22\)

\(4x + 12y = 22\)

Compare with \(x + 3y = 11\).

Multiply \(x + 3y = 11\) by 4 to match the left side.

\(4x + 12y = 44\)

Left sides match but right sides differ (44 vs 22).

So the ratios of coefficients are equal but constants differ ⇒ parallel.

Hence inconsistent.

Idea. For two equations to have infinitely many solutions, the lines must be coincident. This happens when all three ratios are equal:

\(\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2}\).

Step 1: Write in standard form.

First equation: \(\lambda x + 3y = -7\)

Move terms: \(\lambda x + 3y + 7 = 0\)

So \(a_1 = \lambda\), \(b_1 = 3\), \(c_1 = 7\).

Second equation: \(2x + 6y = 14\)

Move terms: \(2x + 6y - 14 = 0\)

So \(a_2 = 2\), \(b_2 = 6\), \(c_2 = -14\).

Step 2: Form the ratios.

\(\dfrac{a_1}{a_2} = \dfrac{\lambda}{2}\)

\(\dfrac{b_1}{b_2} = \dfrac{3}{6} = \dfrac{1}{2}\)

\(\dfrac{c_1}{c_2} = \dfrac{7}{-14} = -\dfrac{1}{2}\)

Step 3: Compare the ratios.

The second ratio is \(\dfrac{1}{2}\), while the third is \(-\dfrac{1}{2}\).

Since one is positive and the other is negative, they cannot be equal, no matter what value of \(\lambda\) is chosen.

Conclusion. The statement is false. There is no value of \(\lambda\) for which the pair of equations has infinitely many solutions.

Step 1: Write equations in standard form.

First equation: \(x - 2y = 8\) ⇒ \(x - 2y - 8 = 0\).

Second equation: \(5x - 10y = c\) ⇒ \(5x - 10y - c = 0\).

So, we have:

\(a_1 = 1,\; b_1 = -2,\; c_1 = -8\)

\(a_2 = 5,\; b_2 = -10,\; c_2 = -c\)

Step 2: Recall the condition for unique solution.

Two linear equations have a unique solution if:

\(\dfrac{a_1}{a_2} \ne \dfrac{b_1}{b_2}\).

Step 3: Compare ratios of coefficients.

\(\dfrac{a_1}{a_2} = \dfrac{1}{5}\).

\(\dfrac{b_1}{b_2} = \dfrac{-2}{-10} = \dfrac{1}{5}\).

The two ratios are equal.

Thus, the lines are either coincident or parallel.

Step 4: Check constants ratio.

\(\dfrac{c_1}{c_2} = \dfrac{-8}{-c} = \dfrac{8}{c}\).

• If \(\dfrac{8}{c} = \dfrac{1}{5}\), then \(c = 40\).

In this case, all three ratios are equal ⇒ the lines are coincident ⇒ infinitely many solutions.

• If \(c \ne 40\), then \(\dfrac{c_1}{c_2} \ne \dfrac{a_1}{a_2}\).

So the lines are parallel and distinct ⇒ no solution.

Step 5: Conclusion.

In neither case do we get a unique solution.

Hence, the given statement “For all real values of \(c\), the pair of equations have a unique solution” is false.

Step 1: Recall the general forms of special lines.

Lines of the form \(x = a\) are vertical lines. They fix the x–coordinate to a constant value and allow the y–coordinate to take any value.

Lines of the form \(y = b\) are horizontal lines. They fix the y–coordinate to a constant value and allow the x–coordinate to vary.

Step 2: Analyze the given line \(x = 7\).

The equation \(x = 7\) means that for every point on this line, the x–coordinate is always 7.

The y–coordinate can take any real value.

Therefore, the line is a vertical line passing through the point \((7, 0)\) on the x–axis.

Step 3: Compare with the coordinate axes.

Vertical lines are parallel to the y–axis, because both extend up and down without changing the x–coordinate.

On the other hand, lines parallel to the x–axis are of the form \(y = b\), which is not the case here.

Conclusion. The line \(x = 7\) is a vertical line, parallel to the y–axis, not the x–axis. Hence, the given statement is false.

Write both in standard form.

\(-x + 2y + 2 = 0\)

\(\dfrac{1}{2}x - \dfrac{1}{4}y + 1 = 0\)

Compare ratios of coefficients.

\(\dfrac{a_1}{a_2} = \dfrac{-1}{\,1/2\,} = -2\)

\(\dfrac{b_1}{b_2} = \dfrac{2}{\,-1/4\,} = -8\)

Since these ratios are unequal, the lines intersect at exactly one point.

Therefore, there is a unique solution.

Rewrite the first equation.

\(4x + 3y - 1 = 5\)

\(4x + 3y = 6\)

The second equation is \(12x + 9y = 15\).

Compute the three ratios.

\(\dfrac{a_1}{a_2} = \dfrac{4}{12} = \dfrac{1}{3}\)

\(\dfrac{b_1}{b_2} = \dfrac{3}{9} = \dfrac{1}{3}\)

\(\dfrac{c_1}{c_2} = \dfrac{6}{15} = \dfrac{2}{5}\)

The constants ratio is different, so the lines are parallel distinct lines.

Hence, they are not coincident.

Compute ratios.

\(\dfrac{a_1}{a_2} = \dfrac{1}{3}\)

\(\dfrac{b_1}{b_2} = \dfrac{2}{6} = \dfrac{1}{3}\)

\(\dfrac{c_1}{c_2} = \dfrac{-3}{-9} = \dfrac{1}{3}\)

All three ratios are equal, so the two equations represent the same line.

Therefore, they are coincident with infinitely many solutions, i.e., consistent.