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February 8, 2016
Posted by **andra ** on Thursday, October 31, 2013 at 5:58am.

y=3x-14

2. x+y=0

5x+y=4

3. 6x+2y=4

4x+2=8

- Substution Soluton MATH -
**Sam**, Thursday, October 31, 2013 at 8:15am1. subtituting we get -5x + 2 = 3x-14

so we get 8x = 16 and x=2, then plugging back in we get y = 2*-5 + 2 = -8

2. subtracting first from second equation gives us 4x = 4 so x = 1, and y = -1

3. 4x + 2 = 8 so 4x = 6, and x = 3/2

also thus 3/2*6 + 2y = 4

2y= -5

y = -5/2

maybe you had a typo and meant the second equation is 4x+2y = 8

so then we subtract frist equation from second one and get 2x = -2, x=-1

then we plug back in for the second one to get -4 + 2y = 8

y= 12/2 = 6

:D

- Substution Soluton MATH -
**Reiny**, Thursday, October 31, 2013 at 8:27am1. since we have two expressions each equal to y, just equate them:

3x-14 = -5x+2

8x = 16

x=2

sub into one of them:

y = 3x-14 = 6-14 = -8

2. from the first : y = -x

into the 2nd:

5x -x = 4

4x=4

x = 1

then y = -1

3. probably a typo , you meant: 4x + 2y = 8

6x+2y=4

4x+2y=8

subtract them:

2x = -4

x = -2

into 4x+2y=8 ---> -8+2y=8

2y = 16

y = 8

- Substution Soluton MATH -
**andra**, Thursday, October 31, 2013 at 8:58amReiny I don't understand the second part of number 3. Please explain. Actually I really didn't understand the whole thing. Why did you subtract the two equations? than when you got the answer for x I don't even know what happened after that

- Substution Soluton MATH -
**Reiny**, Thursday, October 31, 2013 at 9:15am3. Assuming it was a typo and the 2nd equation was 4x+2y = 8,

notice that both terms contain 2y as the y-term.

The method I used is called "elimination".

if we subtract the two equations, of course we can only add/subtract like terms, I get:

6x-4x = 2x

2y - 2y = 0 ----> Ahh, I have "eliminated" the y term

4-8 = -4

giving me

2x + 0 = -4

2x = -4

x = -2

now go back to either of the original equations, I picked the 2nd

4x+2y = 8

4(-2) + 2y = 8

-8 + 2y = 8

2y = 16

y = 8

You should try substituting x=-2 into the first, to see that you get the same answer for y.

I usually pick the easier-looking equation.

after becoming a bit more proficient in your algebra, some of those last steps can be skipped.