Roman C. answered • 09/02/15
Tutor
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(851)
Masters of Education Graduate with Mathematics Expertise
System:
(1): x+3y=7
(2): 6x+21y=51
(2): 6x+21y=51
Solution:
(3)=(2)-6(1): (6x+21y) - 6(x+3y) = 51 - 6·7
6x+21y - 6x - 18y = 51 - 42
3y = 9
y = 3
Now substitute.
x + 3·3 = 7
x + 9 = 7
x = -2
So the solution is x = -2, y = 3.
Roman C.
tutor
The same idea will work with any system of linear equations if the number of unknowns and number of equations is the same.
System:
(1): x + 4y = -1
(2): 12x + 51y = -12
Solution:
(2) - 12(1): (12x + 51y) - 12(x + 4y) = -12 - 12(-1)
12x + 51 y - 12x - 48y = 0
3y = 0
y = 0
Now substitute.
x + 4·0 = -1
x = -1
So the solution is x = -1, y=0.
Report
09/08/15
Katie C.
09/02/15