# rearranging equasions

• May 14th 2010, 06:21 PM
henry
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

• May 14th 2010, 06:35 PM
dwsmith
Quote:

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?
• May 14th 2010, 06:37 PM
Prove It
Quote:

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 $x$ the subject...

$y = 8 - 2x$

$y + 2x = 8$

$2x = 8 - y$

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

$x = 4 - \frac{y}{2}$.
• May 14th 2010, 06:55 PM
henry
i need to find the slope, x intercept and y intercept of both equations
• May 14th 2010, 07:09 PM
henry
Quote:

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
• May 14th 2010, 07:21 PM
Prove It
You can read off the slope and $y$ intercept in the first one, because it is already in $y = mx + c$ form... $m$ is the slope and $c$ is the $y$ intercept...

$y = -2x + 8$.

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

For the second

$3y + 2x = -12$

$3y = -2x - 12$

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

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