solving for substitution

• Aug 10th 2010, 08:49 PM
mafai44
solving for substitution
having trouble understanding substitution this is the problem :

Solve for the system of equations by substitution

5x-7=-y
2x-y=0

This is another one..

6x+7y=1
x=55-9y
• Aug 10th 2010, 08:52 PM
Prove It
$5x-7=-y$
$2x-y=0$.

Substituting $-y$ into the second equation gives

$2x+5x-7=0$

$7x-7=0$

$7x = 7$

$x = 1$.

You know $-y = 5x-7$ so

$-y = 5\cdot 1 - 7$

$-y = 5 - 7$

$-y = -2$

$y = 2$.

Therefore $(x, y) = (1, 2)$ is the solution.
• Aug 10th 2010, 08:54 PM
eumyang
5x-7=-y
2x-y=0

First, solve one of the equations for one of the variables. It really doesn't matter which equation or which variable. You probably will want to choose the way that's the easiest. I'm going to start by solving the 2nd equation for y:

\begin{aligned}
2x - y &= 0 \\
2x &= y \\
\end{aligned}

Now, substitute this into the 1st equation wherever you see a y, and then solve for x:
\begin{aligned}
5x - 7 &= -y \\
5x - 7 &= -2x \\
-7 &= -7x \\
x &= 1
\end{aligned}

Now, plug this into the altered version of the 2nd equation:
\begin{aligned}
2x &= y \\
2(1) &= y \\
y &= 2
\end{aligned}

The answer is (1, 2). You want to try your other example now?

EDIT: Too slow! ;)
• Aug 10th 2010, 08:59 PM
mafai44
thank you! does anyone know solving for elimination?

4x=15+3y
-6/5x+y=-17/5
• Aug 10th 2010, 09:01 PM
Prove It

Is it

$4x=15+3y$
$-\frac{6}{5x}+y = -\frac{17}{5}$?
• Aug 10th 2010, 09:04 PM
mafai44
yes the 6/5x , the x is for both the 6 and 5 not just the 5.. not even sure if that matters but you have it right.
• Aug 10th 2010, 09:12 PM
Prove It
Obviously it's not right if it's meant to be

$4x=15+3y$
$-\frac{6}{5}x+y = -\frac{17}{5}$

since $\frac{6}{5}x = \frac{6x}{5}$, not $\frac{6}{5x}$...

Anyway...

$4x=15+3y$
$-\frac{6}{5}x+y = -\frac{17}{5}$

Multiply the second equation by $3$...

$4x=15+3y$
$-\frac{18}{5}x+3y=-\frac{51}{5}$

$4x-3y=15$
$-\frac{18}{5}x+3y=-\frac{51}{5}$.

$(4x-3y) + \left(-\frac{18}{5}x+3y\right) = 15-\frac{51}{5}$

$\frac{2}{5}x=\frac{24}{5}$

$2x = 24$

$x = 12$.

Substituting into the first equation

$4x=15+3y$

$4\cdot 12 = 15 + 3y$

$48 = 15 + 3y$

$33 = 3y$

$y = 11$.

So $(x, y) = (12, 11)$ is the solution.
• Aug 10th 2010, 09:19 PM
mafai44
Quote:

Originally Posted by eumyang
5x-7=-y
2x-y=0

First, solve one of the equations for one of the variables. It really doesn't matter which equation or which variable. You probably will want to choose the way that's the easiest. I'm going to start by solving the 2nd equation for y:

\begin{aligned}
2x - y &= 0 \\
2x &= y \\
\end{aligned}

Now, substitute this into the 1st equation wherever you see a y, and then solve for x:
\begin{aligned}
5x - 7 &= -y \\
5x - 7 &= -2x \\
-7 &= -7x \\
x &= 1
\end{aligned}

Now, plug this into the altered version of the 2nd equation:
\begin{aligned}
2x &= y \\
2(1) &= y \\
y &= 2
\end{aligned}

The answer is (1, 2). You want to try your other example now?

EDIT: Too slow! ;)

yeah if you can help me out with the other example that would be great.. this is one part of math that i do not like haha
• Aug 10th 2010, 09:20 PM
mafai44
double posted by accident sorry
• Aug 11th 2010, 04:12 AM
eumyang
OP: Prove It's point is that you have to be careful with notation. Ideally, you should learn LaTeX so that it's not ambiguous when you type $\frac{6}{5}x$. 6/5x is really read as $\frac{6}{5x}$. If you are not using LaTeX and you want to indicate the fraction 6/5 times x, use parentheses: (6/5)x.