Question

Solve the system of equations algebraically 3x+2y=4 4x+3y=7 7x+4y+7z=5

Answer 1

First, take the equations 3x+2y=4 and 4x+3y=7 and solve for x. Note, * means to times.

Solving for x
Cancel out y
3x + 2y = 4
4x + 3y = 7

Times equation 1 3x+2y=4 by -3 ad equation 2 by 2
Equation 1
3x + 2y = 4
-3( 3x+2y) = 4 * -3
-9x - 6y = -12

Equation 2:
2(4x+3y) = 7 * 2
8x + 6y = 14

Now we have two new equations and we can cancel out the y. Here are the two new equations:

-9x - 6y = -12
8x + 6y = 14

Lets solve by canceling out the y. -6y + 6y = 0

-9x - 6y = -12
8x + 6y = 14
-----------------------
-1x + 0 = 2
-1x = 2

`\frac{-1x} {-1} = (2)/(-1)`
x = -2

We found x so not lets find y. We can find y like we did for x or we can just insert x = -2 into the equation 3x+2y=4 and solve for y. Lets do that now.
3x+2y=4
3*-2 + 2y = 4
-6 + 2y = 4
-6 + 6 + 2y = 4 + 6
2y = 10

`(2y)/(2) = (10)/(2)`
y = 5

Now that we have found x and y, we can use this new information and insert -2 for x and 5 for y in the equation 7x+4y+7z=5 and solve for z. Lets do that now.
7x + 4y + 7z = 5
7 * -2 + 4 * 5 + 7z = 5
-14 + 20 + 7z = 5
-14 +14 + 20 - 20 + 7z = 5 + 14 - 20
7z = -1

`(7z)/(7) = -(1)/(7)`
z =
`-(1)/(7)`

We are done. So our answers are: x = -2 y = 5 and z =
`-(1)/(7)`