# rearranging equasions

#### henry

Morning, I have not studied maths for some time now and need some help rearranging the 2 following equations.

1. y=8-2x

2. 3y+2x=-12

#### dwsmith

MHF Hall of Honor
Morning, I have not studied maths for some time now and need some help rearranging the 2 following equations.

1. y=8-2x

2. 3y+2x=-12

How are we arranging them?

Are the part of a linear system; therefore, we need to solve them simultaneously?

#### Prove It

MHF Helper
Morning, I have not studied maths for some time now and need some help rearranging the 2 following equations.

1. y=8-2x

2. 3y+2x=-12

I suppose you want to make $$\displaystyle x$$ the subject...

$$\displaystyle y = 8 - 2x$$

$$\displaystyle y + 2x = 8$$

$$\displaystyle 2x = 8 - y$$

$$\displaystyle x = \frac{8 - y}{2}$$

$$\displaystyle x = 4 - \frac{y}{2}$$.

• henry

#### henry

i need to find the slope, x intercept and y intercept of both equations

#### henry

How are we arranging them?

Are the part of a linear system; therefore, we need to solve them simultaneously?

i need to find the slope, x intercept and y intercept of both equations

#### Prove It

MHF Helper
You can read off the slope and $$\displaystyle y$$ intercept in the first one, because it is already in $$\displaystyle y = mx + c$$ form... $$\displaystyle m$$ is the slope and $$\displaystyle c$$ is the $$\displaystyle y$$ intercept...

$$\displaystyle y = -2x + 8$$.

To find the $$\displaystyle x$$ intercept, let $$\displaystyle y = 0$$ and solve for $$\displaystyle x$$.

For the second

$$\displaystyle 3y + 2x = -12$$

$$\displaystyle 3y = -2x - 12$$

$$\displaystyle y = -\frac{2}{3}x - 4$$.

Now you can do the same process as the first question.

• henry