# solve the following system of linear equation by substitution by first isolating $x$?

Here's what I was given:

$x$ + $y$ = 9

-10$x$ + 6$y$ = 6

Firstly, in order to isolate $x$, I changed the first line so that $y$ would equal the equation and I got:

$x$ - 9 = -$y$

In the second equation, I replaced $y$ with $x$ - 9 = -$y$ in order to solve for $x$:

-10$x$ + 6($x$-9) = 6

$x$ = -15

I solved for $y$ by replacing my newfound $x$ value into the equation:

-10(-15) + 6$y$ = 6

$y$ = -24

Final answer: (-15, -24)

(the right answer in the textbook: (3, 6)

What did I do wrong?

Hint: Check whether you substituted the correct value of $y$ in the second equation. $y = 9-x$ not $x-9$.
• @mathisn'tmyforte You wrote $x-9=-y$, which means that $9-x=y$! – Math Lover Aug 8 '17 at 2:30
In the second equation, I replaced $y$ with $x - 9 = -y$ in order to solve for $x$:
This is where you made your error. You cannot replace $y$ with $-y$ since $y \neq -y$ unless $y = 0$.
Let's see how we can fix your error. We are given \begin{align*} x + y & = 9 \tag{1}\\ -10x + 6y & = 6 \tag{2} \end{align*} If we subtract $x$ from both sides of equation 1, we obtain $$y = 9 - x$$ Substituting $9 - x$ for $y$ in equation 2 yields \begin{align*} -10x + 6(9 - x) & = 6 && \text{substitute $9 - x$ for $y$}\\ -10x + 54 - 6x & = 6 && \text{distribute}\\ -16x + 54 & = 6 && \text{combine like terms}\\ -16x & = -48 && \text{subtract $54$ from each side of the equation}\\ x & = 3 && \text{divide both sides of the equation by $-16$} \end{align*} Substituting $3$ for $x$ in the equation $y = 9 - x$ yields $y = 9 - 3 = 6$. Hence, the solution is $(3, 6)$.