PDA

View Full Version : Solving Linear Systems by substitution! HELP!


MathMomma
Jul 18, 2010, 08:09 PM
I CAN NOT FIGURE OUT THIS QUESTION the solution is given as (-1,-5) and I am not getting those two answers!

4(1-2x) = 3(3-y) - 12
2(3x-1)-(y+4) = - 7

KISS
Jul 18, 2010, 09:00 PM
There is a trick that cuts out a LOT of the algebra.

2(3x-1)-(y+4) = - 7

See the -(y+4) --> Make it 3-y

First:
2(3x-1) -4-y = -7

Add 7 to both sides

2(3x-1) + (3-y) = 0

Check original solution since you have it.

2(3(-1)-1) + (3+5) =0; Good, didn't change answer

Solve 2(3x-1) + (3-y) = 0 for 3-y

3-y = -2(3x-1)

Now substitute the -2(3x-1) for 3-y into your first equation.

Who says you have to substitute y?

I know, it's a trick.

KISS
Jul 18, 2010, 09:50 PM
Your # 1 eqn is: 4(1-2x) = 3(3-y) - 12

substituting:

4(1-2x) = 3(-2(3x-1)) - 12

check with -1 for x since we have it.

4(1+2) = 3(-2(-4)) - 12
12 = 24 -12

So, you just have to do this algebraic mess:

4(1-2x) = 3(-2(3x-1)) - 12

to find x.

4-8x = -6(3x-1) - 12

My way:

0 = -6(3x-1) - 12 -4 + 8x

Count x's -18 +8 or -10x
Count #'s: 6 - 16 or -10

0 = -10x-10

10x = -10 ---> x = -1

Substitute into 2(3x-1)-(y+4) = - 7

make it equal to zero; e.g. add 7 to both sides

2(-4) -y - 4 +7 = 0
Count because it's easier.

y's = -1
#'s -8 -4 +7 = -12-7 = -5

-1y - 5 = 0
-y = 5
y = -5

KISS
Jul 18, 2010, 09:55 PM
You can do it the hard way. I can find the mistakes. See a full solution above.

Unknown008
Jul 18, 2010, 10:09 PM
Maybe cleaning the equations from the very start will help?

4(1-2x) = 3(3-y) - 12
2(3x-1)-(y+4) = - 7

These, when expanding become:
4 - 8x = 9 - 3y - 12
6x - 2 - y - 4 = -7

Now, simplify both;
- 8x = -3y -7
6x - y = -1

Hm... too many negatives. I multiply both by -1;
8x = 3y + 7
y - 6x = 1

Now, you can make y the subject of the formula to sub into the first equation, like this:
y = 1 + 6x

8x = 3(1 + 6x) + 7
8x = 3 + 18x + 7

Now, just simplify to get x = -1.

Then, replace that value into " y - 6x = 1 " or " 8x = 3y + 7 " or " y = 1 + 6x " to get the value of y.

There isn't much brackets to worry about, so long as you keep them simple :)

KISS
Jul 18, 2010, 10:18 PM
And, of course there are a few other ways to do it:

The second eqn has y nearly all by itself, so you can solve for y in that eqn. But you can also solve for -y and substitute into the first.

You don't have to find x and then y. You can find y then x.