# Math Help - Help....I need this for monday

1. ## Help....I need this for monday

Hi, I am really stuck with this equations. I am so lost and dont know what to do. Please help.

x - 2 + y - 1 = 13 (eq.1)
3 4 12

2 - x + 3 + y = 11 (eq.2)
2 3 6

2. Originally Posted by NettyJay
Hi, I am really stuck with this equations. I am so lost and dont know what to do. Please help.

x - 2 + y - 1 = 13 (eq.1)
3 4 12

2 - x + 3 + y = 11 (eq.2)
2 3 6
Did you mean:
Originally Posted by NettyJay
Hi, I am really stuck with this equations. I am so lost and dont know what to do. Please help.

$\frac{x - 2}{3} + \frac{y - 1}{4} =
\frac{13}{12}$
(eq.1)

$\frac{2 - x}{2} + \frac{3 + y}{3} = \frac{11}{6}$ (eq.2)

3. Originally Posted by NettyJay
Hi, I am really stuck with this equations. I am so lost and dont know what to do. Please help.

Hi, I am really stuck with this equations. I am so lost and dont know what to do. Please help.

$\frac{x - 2}{3} + \frac{y - 1}{4} = \frac{13}{12}$ (eq.1)

$\frac{2 - x}{2} + \frac{3 + y}{3} = \frac{11}{6}$(eq.2)
I'll do $\frac{x - 2}{3} + \frac{y - 1}{4} = \frac{13}{12}$ you try the other one.

First thing to do is get rid of denominators, so multiply everything by 3 to get: $x - 2 + \frac{3(y - 1)}{4} = \frac{3(13)}{12}$

Now multiply by 4: $4x - 4(2) + 3(y - 1) = \frac{4(39)}{12}$

Simplify: $4x -8 + 3y - 3 = \frac{156}{12}$

Thus: $4x+ 3y - 11 = \frac{156}{12}$

Add 11 to both sides: $4x+ 3y = \frac{156}{12}+11$

Subtract 3y from both sides: $4x = \frac{156}{12}+\frac{12(11)}{12}-3y$

Then divide by 4: $x = \frac{156}{4(12)}+\frac{12(11)}{4(12)}-\frac{3y}{4}$

Solve: $x = \frac{156}{48}+\frac{132}{48}-\frac{12(3y)}{12(4)}=\frac{156+138}{48}-\frac{36y}{48}=\frac{294-36y}{48}$

That can be simplified to: $\boxed{x=\frac{49-6y}{8}}$

4. Was this supposed to be a system of equations, meaning was I sposed to find the value of x and y to fit both of them?

5. Originally Posted by NettyJay
Hi, I am really stuck with this equations. I am so lost and dont know what to do. Please help.

x - 2 + y - 1 = 13 (eq.1)
...3....... 4..... 12

2 - x + 3 + y = 11 (eq.2)
..2........ 3....... 6
Multiply the first equation through by $12$ to get:

$
4x-8+3y-3=13
$

rearrange to:

$
4x + 3y =24
$

Multiply the second equation through by $6$ to get:

$
6-3x+6+2y=11
$

rearrange:

$
-3x+2y=-1
$

So you have a bog standard pair of simultaneous linear equations:

... $
4x + 3y =24
$

$
-3x+2y=-1
$

RonL

6. Just to extend CaptainBlack's post.

You've got system of equations to solve:
$\begin{array}{l}
4x + 3y = 24 \\
- 3x + 2y = - 1 \\
\end{array}
$

Solve first equation for $x$:
$\begin{array}{l}
4x + 3y = 24 \\
4x = 24 - 3y \\
x = \frac{{24 - 3y}}{4} \\
\end{array}
$

Substitute $x$ into second equation:
$\begin{array}{l}
- 3 \cdot (\frac{{24 - 3y}}{4}{\rm{)}} + 2y{\rm{ }} = - 1 \\
- \frac{{72 - 9y}}{4} + 2y = - 1 \\
\frac{{ - 72 + 9y + 8y}}{4} = - 1 \\
\frac{{17y - 72}}{4} = - 1 \\
17y - 72 = - 4 \\
17y = 68 \\
y = 4 \\
\end{array}
$

Now we can find $x$ from first equation by substituting $y$:
$\begin{array}{l}
4x + 3 \cdot 4 = 24 \\
4x + 12 = 24 \\
4x = 12 \\
x = 3 \\
\end{array}
$

So solutions are $x=3,y=4$