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
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
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: $\displaystyle \boxed{x=\frac{49-6y}{8}}$
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
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 $\displaystyle x=3,y=4$