Help....I need this for monday

N

NettyJay

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

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

2 - x + 3 + y = 11 (eq.2)
2 3 6
 
Last edited by a moderator:
May 2006
1,024
134
New England
Hi, I am really stuck with this equations. I am so lost and dont know what to do. Please help.:confused:

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

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

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

\(\displaystyle \frac{2 - x}{2} + \frac{3 + y}{3} = \frac{11}{6}\) (eq.2)
 
May 2006
1,024
134
New England
Hi, I am really stuck with this equations. I am so lost and dont know what to do. Please help.:confused:

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: \(\displaystyle \boxed{x=\frac{49-6y}{8}}\)
 
May 2006
1,024
134
New England
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?
 

CaptainBlack

MHF Hall of Fame
Nov 2005
14,975
5,273
erehwon
Hi, I am really stuck with this equations. I am so lost and dont know what to do. Please help.:confused:

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
 
Mar 2006
340
31
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 \(\displaystyle x=3,y=4\)