Question

3x -4y=11. 7x-5y=4 solve these by substitution and elimination cross multiplication method​

Answer 1

Answer:

x = -3, y = -5

Step-by-step explanation:

By Substitution method

3x - 4y = 11    -------- 1

7x - 5y = 4    -------- 2

3x = 11 + 4y

x = (11 + 4y)/3

Substituting this value of x in equation 2

7x - 5y = 4

(77 + 28y)/3 - 15y/3 = 4

(77 + 28y - 15y) / 3 = 4

77 + 13y = 12

13y = -65

y = -5

3x - 4y = 11

3x + 20 = 11

3x = -9

x = -3

By Elimination Method

3x - 4y = 11    -------- 1

7x - 5y = 4    -------- 2

Multiplying equation 1 by 5, and 2 by 4

15x - 20y = 55    -------- 3

28x - 20y = 16    -------- 4

Subtracting equation 4 from 3

-13x = 39

x = -3

Substituting this value of x in equation 1

-9 - 4y = 11

-4y = 20

y = -5

By Cross Multiplication

$\frac{x}{b_{1}c_{2} - b_{2}c_{1} } = \frac{y}{c_{1}a_{2} - c_{2}a_{1} } = \frac{1}{a_{1}b_{2} - a_{2}b_{1} }$

Here,

$a_{1} = 3; b_{1} = -4; c_{1} = -11\\\\a_{2} = 7; b_{2} = -5; c_{2} = -4$

$\frac{x}{b_{1}c_{2} - b_{2}c_{1} } = \frac{y}{c_{1}a_{2} - c_{2}a_{1} } = \frac{1}{a_{1}b_{2} - a_{2}b_{1} }$

x/-39 = y/-65 = 1/13

x = -3

y = -5

IF YOU FIND IT USEFUL, PLEASE MARK IT AS THE BRAINLIEST ANSWER