Pair of Linear Equations in Two Variables | Exercise 3.2

Question 2

  1. Solve 2x+3y=112x + 3y = 11 and 2x−4y=−242x - 4y = -24 and hence find the value of 'm' for which y=mx+3y = mx + 3.
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Solution

Let's solve the system of linear equations to find the values of xx and yy. Then we will use these values to find 'm'.

Step 1 — Solve for x and y

We have two equations. Let's call them equation (1) and equation (2).

2x+3y=11(1)2x + 3y = 11 \quad \text{(1)}

2x−4y=−24(2)2x - 4y = -24 \quad \text{(2)}

From equation (1), we can express yy in terms of xx. We will isolate yy.

3y=11−2x3y = 11 - 2x

y=11−2x3y = \frac{11 - 2x}{3}

Now, we substitute this value of yy into equation (2). This helps us find the value of xx.

2x−4(11−2x3)=−242x - 4\left(\frac{11 - 2x}{3}\right) = -24

Let's multiply the entire equation by 3. This removes the fraction.

3(2x)−4(11−2x)=3(−24)3(2x) - 4(11 - 2x) = 3(-24)

6x−44+8x=−726x - 44 + 8x = -72

Combine the terms with xx.

14x−44=−7214x - 44 = -72

Add 44 to both sides.

14x=−72+4414x = -72 + 44

14x=−2814x = -28

Divide by 14 to find xx.

x=−2814x = \frac{-28}{14}

x=−2\boxed{x = -2}

Now we substitute x=−2x = \mathbf{-2} back into the expression for yy. We will use y=11−2x3y = \frac{11 - 2x}{3}.

y=11−2(−2)3y = \frac{11 - 2(-2)}{3}

y=11+43y = \frac{11 + 4}{3}

y=153y = \frac{15}{3}

y=5\boxed{y = 5}

Step 2 — Find m

We are given the equation y=mx+3y = mx + 3. We found x=−2x = \mathbf{-2} and y=5y = \mathbf{5}. Let's substitute these values into the equation.

5=m(−2)+35 = m(-2) + 3

5=−2m+35 = -2m + 3

Subtract 3 from both sides.

5−3=−2m5 - 3 = -2m

2=−2m2 = -2m

Divide by -2 to find mm.

m=2−2m = \frac{2}{-2}

m=−1\boxed{m = -1}

Answer

(i) The value of mm is -1.

More questions in Exercise 3.2

Q1
  1. Solve the following pair of linear equations by the substitution method.

(i) x+y=14x + y = 14
x−y=4x - y = 4

(ii) s−t=3s - t = 3
s3+t2=6\frac{s}{3} + \frac{t}{2} = 6

(iii) 3x−y=33x - y = 3
9x−3y=99x - 3y = 9

(iv) 0.2x+0.3y=1.30.2x + 0.3y = 1.3
0.4x+0.5y=2.30.4x + 0.5y = 2.3

(v) 2x+3y=0\sqrt{2}x + \sqrt{3}y = 0
3x−8y=0\sqrt{3}x - \sqrt{8}y = 0

(vi) 3x2−5y3=−2\frac{3x}{2} - \frac{5y}{3} = -2
x3+y2=136\frac{x}{3} + \frac{y}{2} = \frac{13}{6}

Q2
  1. Solve 2x+3y=112x + 3y = 11 and 2x−4y=−242x - 4y = -24 and hence find the value of 'm' for which y=mx+3y = mx + 3.
Q3
  1. Form the pair of linear equations for the following problems and find their solution by substitution method.

(i) The difference between two numbers is 26 and one number is three times the other. Find them.

(ii) The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.

(iii) The coach of a cricket team buys 7 bats and 6 balls for ₹ 3800. Later, she buys 3 bats and 5 balls for ₹ 1750. Find the cost of each bat and each ball.

(iv) The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of 10 km, the charge paid is ₹ 105 and for a journey of 15 km, the charge paid is ₹ 155. What are the fixed charges and the charge per km? How much does a person have to pay for travelling a distance of 25 km?

(v) A fraction becomes 911\frac{9}{11}, if 2 is added to both the numerator and the denominator. If, 3 is added to both the numerator and the denominator it becomes 56\frac{5}{6}. Find the fraction.

(vi) Five years hence, the age of Jacob will be three times that of his son. Five years ago, Jacob's age was seven times that of his son. What are their present ages?

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