# substitution

• Jun 2nd 2007, 11:52 PM
ncbabe
substitution
alright i know how you substitute when they give you something easy like
6x+7=5
y=5+3

But when you have something like
9x-7y=5
5x+3y=-1 i don't understand you you can make the y= or x=

thank you
• Jun 3rd 2007, 12:14 AM
tukeywilliams
So you have:

$9x-7y = 5$
$5x + 3y = -1$

So first decide what variable you want to solve for in terms of the other variable. I want to solve for $x$ in terms of $y$. So I look at the first equation:

$9x-7y = 5$
$9x = 5+7y$

$x = \frac{5+7y}{9}$

So now I have $x$ in terms of $y$.

Now go to the second equation and substitute in this expression.

$5\left(\frac{5+7y}{9}\right ) + 3y = -1$. You can then solve for $y$. After solving for $y$ plug this value back into the first equation and solve for $x$.

So you get $x = \; \text{something}, \; y = \; \text{something}$
• Jun 3rd 2007, 12:15 AM
earboth
Quote:

Originally Posted by ncbabe
...
But when you have something like
9x-7y=5
5x+3y=-1 i don't understand you you can make the y= or x=
...

Hello,

1. You have to determine which variable should be substituted. In this example I will substitute the y of the first equation by the y of the second equation.

2. Thus we need the y of the second equation in explicite(?) form:

$5x+3y=-1 \Longleftrightarrow 3y=-5x-1 \Longleftrightarrow y=-\frac{5}{3}x-\frac{1}{3}$

3. Now substitute the y of the first equation by the term for y:

$9x-7 \cdot \left(-\frac{5}{3}x-\frac{1}{3} \right)=5 \Longleftrightarrow 9x+\frac{35}{3}x+\frac{7}{3}=5$

4. Collect like terms: x-values at the LHS and all constant values at the RHS:
$\frac{62}{3}x=\frac{8}{3}$

5. Divide by the coefficient of x:

$x=\frac{\frac{8}{3}}{\frac{62}{3}} = \frac{4}{31}$

EDIT:
6. Claculate the y-value by substituting the x-value into the equation: $y=-\frac{5}{3}x-\frac{1}{3}$

$y=-\frac{5}{3}\cdot \frac{4}{31}-\frac{1}{3} = -\frac{17}{31}$