System 1: The solution is (x, y) = (-4, 5)
System 2: The solution is [tex](x, y) = (\frac{11}{3}, -3)[/tex]
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)
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
[tex]x = \frac{11}{3}[/tex]
Substitute the above value of x in eqn 3
y = 8 - 3x
[tex]y = 8 - 3 \times \frac{11}{3}\\\\y = 8 - 11\\\\y = -3[/tex]
Thus the solution is [tex](x, y) = (\frac{11}{3}, -3)[/tex]
