Answer: (1) x = 5 , y = 4
(2) x = -3 and y = 0
(3) x = 2 and y = 2
Explanation:
(1) 2x - 3y = -2 ................... equation 1
2x + y = 14 ................ equation 2
solving the system of linear equation by elimination method. We need to decide the variable to eliminate first , in this case , since the coefficient of x are the same and they have the same signs (+), we can eliminate the variable x first by subtracting equation 1 from equation 2, so we have
2x - 2x + y - (-3y ) = 14 - ( - 2)
4y = 16
divide through by 4
y = 4
substitute y = 4 into equation 1 , we have
2x - 3 (4) = -2
2x - 12 = -2
2x = -2 + 12
2x = 10
x = 5
Therefore :
x = 5 and y = 4
(2) x = -6y - 3 ....................... equation 1
8x + 8y = -24 ....................... equation 2
Solving the system of linear equation by substitution method , substitute x = -6y - 3 into equation 2 , equation 2 becomes
8(-6y - 3 ) + 8y = -24
expanding , we have
-48y - 24 + 8y = -24
-40y - 24 = -24
Add 24 to both sides , we have
- 40y = -24 + 24
-40y = 0
divide through by -40
y = 0/-40
y = 0
substitute y = 0 into equation 1 , equation 1 then becomes
x = -6(0) - 3
x = -3
Therefore : x = -3 and y = 0
(3) 5x + 5y = 20 ........................ equation 1
-3x + 5y = 4 ............................ equation 2
solving the system of linear equation by elimination method , we have to decide the variable to eliminate first , since the coefficient of y are the same and are both positive, we will eliminate y by subtracting equation 2 from equation 1 , we have
5x - (-3x) + 5y - 5y = 20 - 4
5x + 3x + 0 = 16
8x = 16
x = 2
substitute x = 2 into equation 1 , equation becomes
5(2) + 5y = 20
10 + 5y = 20
5y = 20 - 10
5y = 10
y = 2
Therefore : x = 2 and y = 2