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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