Solve each system by substitution?

Question

a)

y = -3x + 3

7x – 2y = 20

b)

3x + y = 3

x + y = 5

c)

y = x + 1

y = -x + 6

d)

x – y + 3 = 0

x + 2y = 3

Answer 1

a) 7x - 2(-3x+3) = 20

7x - -6x - 6 = 20

7x + 6x = 26

13x = 26

x = 2

Plug in x to find y:

y = -3(2) + 3

y = -6 + 3

y = -3

Check:

7(2) - 2(-3) = 20

14 - -6 = 20, so it checks. x = 2, y = -3

b) 3x + (5-x) = 3

2x + 5 = 3

2x = -2

x = -1

-1 + y = 5

y = 6

Check: 3(-1) + 6 = 3

-3 + 6 = 3, so it checks. x = -1, y = 6

c) -x + 6 = x + 1

6 = 2x + 1

5 = 2x

x = 2.5

y = 2.5 + 1

y = 3.5

Check: 3.5 = -2.5 + 6, so it checks. x = 2.5, y = 3.5

d) y = (3-x)/2

x = y - 3

y = (3 - (y-3))/2

y = (3 - y + 3)/2

y = (6-y)/2

2y = 6-y

3y = 6

y = 2

x - 2 + 3 = 0

x = -1

Check: -1 + (2*2) = 3

-1 + 4 = 3, so it checks. x = -1, y = 2

=)

Answer 2

y = -3x + 3

7x – 2y = 20

7x-2(-3x+3)=20

7x+6x-3=20

11x=17

x=17/11

y= -3(17/11)+3

y=-51/11 +33/11

y=-18/11

x=17/11 y=-18/11

3x+y=3

x+y=5

y=5-x

3x+5-x=3

2x+5=3

2x=-2

x=-1

x+y=5

-1+y=5

y=6

y=x+1

y=-x+6

x+1=-x+6

2x+1=6

2x=5

x=5/2

x-y+3=0

x+2y=3

x=3-2y

3-2y-y+3=0

-3y=0

y=0

x+2y=3

x+2(0)=3

x=3

Answer 3

They have pretty much given you the answers, but haven't explained what they did, so I will.

In substitution to solve two equations, you do exactly what it says... you substitute the first for part of the second.  For instance...

y =  -3 + 3 is easy because y is isolated on one side of the equation

So... plug y into the second equation...

7x - 2(-3x + 3) = 20

7x + 6x -6 = 20

13x - 6 = 20

13x - 6 + 6 = 20 + 6  (what you do to one side of an equation you have to do to the other side to keep it equal)

So 13x = 26

divide both sides by 13 to reduce and you get x = 2

Simple, huh?  Now, the trick is... this procedure works for all problems like this.  Just remember that you have to isolate the unknown in one of the problems... in other words, get it by itself on one side of the equation... and then substitute it into the other problem.

The second one would be easier if you isolated the second problem...

x + y = 5

x + y - y = 5 - y  (remember, what you do to one side....... )

so x = 5 - y

Now plug that into the first problem...

3 (5 - y) + y = 3

15 - 3y + y = 3

15 - 2y = 3

15 - 3 - 2y + 2y = 3 - 3 + 2y  (this gets the y on one side and the numerical term on the other side.  What I did to one side, I did to the other, it's just that I did two things at the same time.  Look at it closely to see what I did.)

12 = 2y

6 = y

Hope this helps you solve future problems.

:)