1. ## rearranging equasions

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

2. Originally Posted by 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

How are we arranging them?

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

3. Originally Posted by 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

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

$\displaystyle y = 8 - 2x$

$\displaystyle y + 2x = 8$

$\displaystyle 2x = 8 - y$

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

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

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

5. Originally Posted by dwsmith
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

6. 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.