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
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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
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.
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
You can do it the hard way. I can find the mistakes. See a full solution above.
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 :)
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.
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