# Thread: 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.

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

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

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

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

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

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

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

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

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

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

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

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

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

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 $\displaystyle 12$ to get:

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

rearrange to:

$\displaystyle 4x + 3y =24$

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

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

rearrange:

$\displaystyle -3x+2y=-1$

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

...$\displaystyle 4x + 3y =24$
$\displaystyle -3x+2y=-1$

RonL

6. Just to extend CaptainBlack's post.

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

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

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

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

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