Substitution
y = 4x
x + y = 5
( 1, 4)
x – y = 1
x+y=3
(2, 1)
x + 4y = 11
x – 6y = 11
(11, 0)
2x – 3y = 21
5x – 2y = 25
(3, -5)
5x + 3y = 16
3x – 5y = –4
elimination using multiplication because none of the variables have a coefficient of 1 or -1 and they cannot be eliminated using addition/subtraction
x = –4y
3x + 2y = 20
( 8, -2)
–x + y = 1
x + y = 11
(5 ,6)
3x + 4y = 19
3x + 6y = 33
(-3 ,7)
5x – 2y = –10
3x + 6y = 66
(2, 10)
3x – 5y = 7
2x + 5y = 13
elimination using addition because y has opposite coefficients in two equations.
x + 2y = 13
3x – 5y = 6
(7, 3)
3x + 4y = 2
4x – 4y = 12
(2, -1)
2x – 3y = 9
–5x – 3y = 30
(-3, -5)
7x + 4y = –4
5x + 8y = 28
(-4, 6)
6x – y = –145
x = 4 – 2y
substitution because x in the second equation has a coefficient of 1.
3x – y = 4
2x – 3y = –9
(3, 5)
5x – y = –6
–x + y = 2
(-1, 1)
6x – 2y = 32
4x – 2y = 18
( 25, 41)
5x + 3y = –10
3x + 5y = –6
(-2, 0)
14x + 7y = 217
14x + 3y = 189
2.5x + y = –2
3x + 2y = 0
( -2, 3)
3x + 2y = –19
–3x – 5y = 25
(-5 ,-2)
7x + 4y = 2
7x + 2y = 8
(2 , -3)
2x + 3y = 14
3x – 4y = 4
(4, 2)
y = 3x – 24
5x – y = 8
substitution because y has a coefficient of 1