rearranging equasions

May 2010
6
0
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

Your help is appreciated
 

dwsmith

MHF Hall of Honor
Mar 2010
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Florida
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

Your help is appreciated
How are we arranging them?

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

Prove It

MHF Helper
Aug 2008
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4,999
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

Your help is appreciated
I suppose you want to make \(\displaystyle x\) the subject...

\(\displaystyle y = 8 - 2x\)

\(\displaystyle y + 2x = 8\)

\(\displaystyle 2x = 8 - y\)

\(\displaystyle x = \frac{8 - y}{2}\)

\(\displaystyle x = 4 - \frac{y}{2}\).
 
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May 2010
6
0
i need to find the slope, x intercept and y intercept of both equations
 
May 2010
6
0
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
Aug 2008
12,883
4,999
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.
 
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