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
$\displaystyle 5x-7=-y$
$\displaystyle 2x-y=0$.
Substituting $\displaystyle -y$ into the second equation gives
$\displaystyle 2x+5x-7=0$
$\displaystyle 7x-7=0$
$\displaystyle 7x = 7$
$\displaystyle x = 1$.
You know $\displaystyle -y = 5x-7$ so
$\displaystyle -y = 5\cdot 1 - 7$
$\displaystyle -y = 5 - 7$
$\displaystyle -y = -2$
$\displaystyle y = 2$.
Therefore $\displaystyle (x, y) = (1, 2)$ is the solution.
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:
$\displaystyle \begin{aligned}
2x - y &= 0 \\
2x &= y \\
\end{aligned}$
Now, substitute this into the 1st equation wherever you see a y, and then solve for x:
$\displaystyle \begin{aligned}
5x - 7 &= -y \\
5x - 7 &= -2x \\
-7 &= -7x \\
x &= 1
\end{aligned}$
Now, plug this into the altered version of the 2nd equation:
$\displaystyle \begin{aligned}
2x &= y \\
2(1) &= y \\
y &= 2
\end{aligned}$
The answer is (1, 2). You want to try your other example now?
EDIT: Too slow!
Obviously it's not right if it's meant to be
$\displaystyle 4x=15+3y$
$\displaystyle -\frac{6}{5}x+y = -\frac{17}{5}$
since $\displaystyle \frac{6}{5}x = \frac{6x}{5}$, not $\displaystyle \frac{6}{5x}$...
Anyway...
$\displaystyle 4x=15+3y$
$\displaystyle -\frac{6}{5}x+y = -\frac{17}{5}$
Multiply the second equation by $\displaystyle 3$...
$\displaystyle 4x=15+3y$
$\displaystyle -\frac{18}{5}x+3y=-\frac{51}{5}$
$\displaystyle 4x-3y=15$
$\displaystyle -\frac{18}{5}x+3y=-\frac{51}{5}$.
Now add the equations together
$\displaystyle (4x-3y) + \left(-\frac{18}{5}x+3y\right) = 15-\frac{51}{5}$
$\displaystyle \frac{2}{5}x=\frac{24}{5}$
$\displaystyle 2x = 24$
$\displaystyle x = 12$.
Substituting into the first equation
$\displaystyle 4x=15+3y$
$\displaystyle 4\cdot 12 = 15 + 3y$
$\displaystyle 48 = 15 + 3y$
$\displaystyle 33 = 3y$
$\displaystyle y = 11$.
So $\displaystyle (x, y) = (12, 11)$ is the solution.
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 $\displaystyle \frac{6}{5}x$. 6/5x is really read as $\displaystyle \frac{6}{5x}$. If you are not using LaTeX and you want to indicate the fraction 6/5 times x, use parentheses: (6/5)x.