Question

CL 6-131. Solve each system using the method of your choice.

Answer 1

System 1: The solution is (x, y) = (-4, 5)

System 2:  The solution is $(x, y) = (\frac{11}{3}, -3)$

Solution:

Given system of equations are:

2x + 3y = 7 ------ eqn 1

-3x - 5y = -13 --------- eqn 2

We can solve by elimination method

Multiply eqn 1 by 3

6x + 9y = 21 ------ eqn 3

Multiply eqn 2 by 2

-6x - 10y = -26 ------- eqn 4

Add eqn 3 and eqn 4

6x + 9y -6x - 10y = 21 - 26

-y = -5

y = 5

Substitute y = 5 in eqn 1

2x + 3(5) = 7

2x + 15 = 7

2x = -8

x = -4

Thus the solution is (x, y) = (-4, 5)

Second system of equation is:

8 - y = 3x ------ eqn 1

2y + 3x = 5 ----- eqn 2

We can solve by susbtitution method

From given,

y = 8 - 3x ----- eqn 3

Substitute eqn 3 in eqn 2

2(8 - 3x) + 3x = 5

16 - 6x + 3x = 5

3x = 16 - 5

3x = 11

$x = \frac{11}{3}$

Substitute the above value of x in eqn 3

y = 8 - 3x

$y = 8 - 3 \times \frac{11}{3}\\\\y = 8 - 11\\\\y = -3$

Thus the solution is $(x, y) = (\frac{11}{3}, -3)$