Algebra Substitution

Question

3 problems -

Problem 1

y=3x

x+y= -32

Problem 2

y=2x+7

y=x-1

Problem 3

x= -2y+4

2x+4y=8

Answer 1

I'll do the first one and you can do the rest:

Problem 1

y=3x

x+y= -32

This is two equations and two unknowns so there are at most one solution (x,y):

Label the first equation as Equation [1] and the second as Equation [2]:

[1] y=3x

[2] x+y= -32

If we substitute y=3x from [1] into [2] we eliminate y:

[2] x + (3x)=-32

4x=-32

x=-8

Now inserting x=-8 into [1]:

y=3(-8)=-24

Solution (x,y)=(-8,-24)

Answer 2

Whenever you have x on one side of an equation and y on the other side, you can substitute one for the other. For example, in problem 2, you have y=2x+7 and y = x -1. Since y = x - 1, you can substitute x - 1 for y in the first equation like so: x - 1 = 2x - 7. Now, you can combine the x's on one side and the numbers on the other:

x - x - 1 + 7 = 2x - x - 7 + 7

Now, do the addition and subtraction and you get

-1 + 7 = x + 0

6 = x.

Now, you should be able to solve problem 3 the same way: substitute "-2y + 4" for x in "2x + 4y = 8"