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

Exercise 3.1

Step 1: Write equations in standard form.

Equation 1: \(6x - 3y + 10 = 0\) ⇒ coefficients \(a_1 = 6\), \(b_1 = -3\), \(c_1 = 10\).

Equation 2: \(2x - y + 9 = 0\) ⇒ coefficients \(a_2 = 2\), \(b_2 = -1\), \(c_2 = 9\).

Step 2: Compare the ratios of coefficients.

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

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

\(\dfrac{c_1}{c_2} = \dfrac{10}{9}\)

Step 3: Apply the condition for parallel lines.

If \(\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} \neq \dfrac{c_1}{c_2}\), then the lines are parallel and distinct.

Here, \(\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = 3\), but \(\dfrac{c_1}{c_2} = \dfrac{10}{9}\), which is not equal to 3.

Conclusion: The two lines are parallel distinct lines.

Step 1: Write coefficients.

From the first equation \(x + 2y + 5 = 0\):

\(a_1 = 1,\; b_1 = 2,\; c_1 = 5\).

From the second equation \(-3x - 6y + 1 = 0\):

\(a_2 = -3,\; b_2 = -6,\; c_2 = 1\).

Step 2: Check ratios of coefficients.

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

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

So, \(\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2}\).

Step 3: Check constant ratio.

\(\dfrac{c_1}{c_2} = \dfrac{5}{1} = 5\).

This is not equal to \(-\dfrac{1}{3}\).

Step 4: Interpret.

Since the first two ratios are equal but different from the third, the lines are parallel and distinct.

Hence, the pair of equations has no solution.

Step 1: A system of linear equations is called consistent if it has at least one solution.

Step 2: For two lines in a plane, there are three possibilities:

• The lines intersect at a single point ⇒ one solution.
• The lines are coincident ⇒ infinitely many solutions.
• The lines are parallel and distinct ⇒ no solution.

Step 3: Since consistency means the equations have one or more solutions, the possible cases are:

• Lines intersecting at one point, or

• Lines coincident (overlapping).

Therefore: If a pair of linear equations is consistent, the lines will be intersecting or coincident.

Step 1: Consider the first equation \(y = 0\). This represents the x-axis, a horizontal line through the origin.

Step 2: The second equation is \(y = -7\). This represents another horizontal line, parallel to the x-axis, but 7 units below it.

Step 3: Two horizontal lines that are distinct will never meet, because they are parallel.

Conclusion: Since the lines do not intersect at any point, the pair of equations has no solution.

Step 1: Understand the equations.

The equation \(x = a\) is a vertical line passing through all points where the \(x\)-coordinate is always \(a\).

The equation \(y = b\) is a horizontal line passing through all points where the \(y\)-coordinate is always \(b\).

Step 2: Check their intersection.

The vertical line \(x = a\) and the horizontal line \(y = b\) meet at a single point.

This point has coordinates where \(x = a\) and \(y = b\), i.e., the point \((a, b)\).

Step 3: Verify the options.

(A) Parallel → Incorrect, since they cross each other.

(B) Intersecting at \((b,a)\) → Incorrect, order of coordinates is wrong.

(C) Coincident → Incorrect, because the lines are distinct and not overlapping.

(D) Intersecting at \((a,b)\) → Correct, since that is the only common point.

Final Answer: Option (D), the lines intersect at \((a,b)\).

Step 1: Write the equations in standard form.

First equation: \(3x - y + 8 = 0\)

Here \(a_1 = 3\), \(b_1 = -1\), \(c_1 = 8\).

Second equation: \(6x - ky = -16\)

Bring RHS to LHS: \(6x - ky + 16 = 0\)

Here \(a_2 = 6\), \(b_2 = -k\), \(c_2 = 16\).

Step 2: Condition for coincident lines.

For two lines to be coincident:

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

Step 3: Compute the ratios.

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

\(\dfrac{c_1}{c_2} = \dfrac{8}{16} = \dfrac{1}{2}\)

So, \(\dfrac{b_1}{b_2}\) must also equal \(\dfrac{1}{2}\).

Step 4: Solve for \(k\).

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

Equating: \(\dfrac{1}{k} = \dfrac{1}{2}\)

\(\Rightarrow k = 2\)

Final Answer: The lines are coincident when \(k = 2\).

Step 1: Recall the condition for parallel lines.

Two lines \(a_1x + b_1y + c_1 = 0\) and \(a_2x + b_2y + c_2 = 0\) are parallel if

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

Step 2: Write equations in standard form.

First line: \(3x + 2ky = 2\)

Rewrite: \(3x + 2ky - 2 = 0\)

So, \(a_1 = 3, \; b_1 = 2k, \; c_1 = -2\).

Second line: \(2x + 5y + 1 = 0\)

So, \(a_2 = 2, \; b_2 = 5, \; c_2 = 1\).

Step 3: Apply parallel condition.

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

That is: \(\dfrac{3}{2} = \dfrac{2k}{5}\).

Step 4: Solve for \(k\).

Cross multiply: \(3 \times 5 = 2 \times 2k\).

\(15 = 4k\).

\(k = \dfrac{15}{4}\).

Step 5: Conclusion.

Therefore, the value of \(k\) is \(\dfrac{15}{4}\).

Correct Option: (C)

Step 1: Write equations in standard form.

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

So, \(a_1 = c, \; b_1 = -1, \; c_1 = -2\).

Second equation: \(6x - 2y = 3\) ⇒ \(6x - 2y - 3 = 0\).

So, \(a_2 = 6, \; b_2 = -2, \; c_2 = -3\).

Step 2: Condition for infinitely many solutions.

We need

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

Step 3: Compare ratios.

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

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

Step 4: Observe.

Since \(\dfrac{1}{2} \ne \dfrac{2}{3}\), the second and third ratios are not equal.

This inequality is true regardless of the value of \(c\). Hence, it is impossible to satisfy the condition for infinitely many solutions.

Conclusion. There is no value of \(c\) that makes the system have infinitely many solutions.

Step 1: Recall condition for dependent equations.

If two linear equations are dependent, one is just a multiple of the other. That means all their coefficients are in the same ratio.

Step 2: Start with the given equation.

Equation: \(-5x + 7y = 2\).

Step 3: Multiply to create an equivalent equation.

If we multiply the entire equation by \(-2\):

\((-2)\times(-5x) + (-2)\times(7y) = (-2)\times 2\)

\(10x - 14y = -4\)

Step 4: Compare with the options.

This matches exactly with Option D.

Final Answer: Option D (\(10x - 14y = -4\)).

Step 1: Check option (A).

Equation 1: \(x + y = -1\). Substituting \(x = 2, y = -3\):

\(2 + (-3) = -1\) ✓ satisfied.

Equation 2: \(2x - 3y = -5\). Substituting:

\(2(2) - 3(-3) = 4 + 9 = 13 \ne -5\).

So, option (A) is not correct.

Step 2: Check option (B).

Equation 1: \(2x + 5y = -11\). Substituting:

\(2(2) + 5(-3) = 4 - 15 = -11\) ✓ satisfied.

Equation 2: \(4x + 10y = -22\). Substituting:

\(4(2) + 10(-3) = 8 - 30 = -22\) ✓ satisfied.

But notice: the second equation is exactly 2 × (first equation). So both equations represent the same line ⇒ infinitely many solutions, not a unique one.

Step 3: Check option (C).

Equation 1: \(2x - y = 1\). Substituting:

\(2(2) - (-3) = 4 + 3 = 7 \ne 1\).

So, option (C) is not correct.

Step 4: Check option (D).

Equation 1: \(x - 4y - 14 = 0\). Substituting:

\(2 - 4(-3) - 14 = 2 + 12 - 14 = 0\) ✓ satisfied.

Equation 2: \(5x - y - 13 = 0\). Substituting:

\(5(2) - (-3) - 13 = 10 + 3 - 13 = 0\) ✓ satisfied.

Both equations are satisfied and they are not multiples of each other ⇒ two distinct lines intersecting at exactly one point.

Conclusion. Option (D) gives a pair of equations with a unique solution at \((2, -3)\).

Step 1: We are given the two equations:

\(x - y = 2\)   ...(i)

\(x + y = 4\)   ...(ii)

Step 2: Add equations (i) and (ii):

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

\(2x = 6\)

\(x = 3\)

Step 3: Substitute \(x = 3\) into equation (i):

\(3 - y = 2\)

\(-y = 2 - 3 = -1\)

\(y = 1\)

Step 4: Thus, the solution is:

\(a = 3\), \(b = 1\)

Final Answer: Option (C) — 3 and 1.

Step 1: Define variables.

Let the number of Re 1 coins be \(r\).

Let the number of Rs 2 coins be \(t\).

Step 2: Form equations from the conditions.

Total number of coins is 50:

\(r + t = 50\).

Total value of coins is Rs 75:

\(1 \times r + 2 \times t = 75\).

So the second equation is \(r + 2t = 75\).

Step 3: Solve the system of equations.

Subtract the first equation from the second:

\((r + 2t) - (r + t) = 75 - 50\)

\(t = 25\).

Now substitute \(t = 25\) in the first equation:

\(r + 25 = 50\)

\(r = 25\).

Step 4: Final Answer.

The number of Re 1 coins = 25, and the number of Rs 2 coins = 25.

So the correct option is (D) 25 and 25.

Step 1: Define the variables.

Let the son's present age be \(s\) years. Then the father's present age is \(f\) years.

According to the problem, the father is six times the son's age:

\(f = 6s\)

Step 2: Use the condition about their ages after 4 years.

After 4 years, the son's age will be \(s + 4\).

The father's age will be \(f + 4\).

It is given that at that time the father’s age will be four times the son’s age:

\(f + 4 = 4(s + 4)\)

Step 3: Substitute \(f = 6s\) into the equation.

\(6s + 4 = 4(s + 4)\)

Step 4: Simplify.

\(6s + 4 = 4s + 16\)

\(6s - 4s = 16 - 4\)

\(2s = 12\)

\(s = 6\)

Step 5: Find the father's age.

Since \(f = 6s\),

\(f = 6 \times 6 = 36\)

Final Answer:

The son's present age is 6 years, and the father's present age is 36 years.