Postby **jg.allinsymbols** » Fri Feb 15, 2013 8:18 am

The form in which you want to use your equations is often your choice, but your notation needs improvement. One would assume you are dealing with linear equations as found in beginning and intermediate algebra. YOUR given equations are probably these:

y=(1/4)x +7, y=-(3/5)x - 4, using parentheses for clarity.

You could arrange the equations so the variable terms come first, and the constant terms are on the other side of equality.

(1/4)x-y=7

AND

-(3/5)x-y=4

Or, better,

(1/4)x-y=7

AND

(3/5)x+y=-4

Better comfort in solving if we clear away the fractions. Multiply the "7" equation by 4 and multiply the "-4" equation by 5 to get:

A Good System Now:

----------------------------

x-4y=28

AND

3x+5y=-20

-----------------------------

You wanted to solve by elimination. First, we can multiply the first equation by 5, and multiply the second equation by 4. This will allow us to eliminate terms of y and solve for x:

5(x-4y)=5*28

and

4(3x+5y)=4*(-20)

5x-20y=140

and

12x+20y=-80

We ADD those two resulting equations giving us this:

5x-20y+12x+20y=140+(-80)

17x=60

**x=60/17**, ONE RESULT FOUND.

Now we can do similarly to eliminate x. We start again from here:

x-4y=28

AND

3x+5y=-20

Multiply the first equation by 3, and then subtract one equation from the other.

3x-12y=84

AND

3x+5y=-20

Try subtracting first from the second:

3x+5y-3x--12y=-20-84

5y+12y=-104

17y=-104

**y=-104/17**, OUR OTHER RESULT FOUND