View Full Version : help!!
kizzy101
08-29-2005, 07:44 PM
i forgot how to do problems like this :( , and it has to be solved by substitution:
8x - 2y = 58
6x - 2y = 40
if someone would help i will be forever greatful :D !
8x - 2y = 58
6x - 2y = 40
Two methods exist to solve these problems
subtract the 2 equations and the y variable will cancel out.
8x-2y-(6x-2y) =58-40
8x-2y-6x+2y =18
2x=18
x=9
sub this back into either of the original equations and you will be done.
soroban
08-30-2005, 12:24 PM
Hello, kizzy101!
Solve by substitution:
[1] . 8x - 2y = 58
[2] . 6x - 2y = 40
(1) Solve one of the equations for one of its variables.
. . . Solve [1] for y: .-2y .= .-8x + 58 . ---> . y .= .4x - 29 . (a)
(2) Substitute into the other equation.
. . . 6x - 2(4x - 29) .= .40
. . . . . 6x - 8x + 58 .= .40
. . . . . . . . . . . . -2x .= .-18
. . . . . . . . . . . . . .x .= .9
Substitute into (a): . y .= .4(9) - 29 .= .7
Answer: . (x, y) .= .(9, 7)
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