Consider the first equation. Subtract $2y$ from both sides.
$$x+15-2y=0$$
Subtract $15$ from both sides. Anything subtracted from zero gives its negation.
$$x-2y=-15$$
Consider the second equation. Subtract $y$ from both sides.
$$x-y=10$$
To solve a pair of equations using substitution, first solve one of the equations for one of the variables. Then substitute the result for that variable in the other equation.
$$x-2y=-15,x-y=10$$
Choose one of the equations and solve it for $x$ by isolating $x$ on the left hand side of the equal sign.
$$x-2y=-15$$
Add $2y$ to both sides of the equation.
$$x=2y-15$$
Substitute $2y-15$ for $x$ in the other equation, $x-y=10$.
$$2y-15-y=10$$
Add $2y$ to $-y$.
$$y-15=10$$
Add $15$ to both sides of the equation.
$$y=25$$
Substitute $25$ for $y$ in $x=2y-15$. Because the resulting equation contains only one variable, you can solve for $x$ directly.
$$x=2\times 25-15$$
Multiply $2$ times $25$.
$$x=50-15$$
Add $-15$ to $50$.
$$x=35$$
The system is now solved.
$$x=35,y=25$$
Steps Using Matrices
Consider the first equation. Subtract $2y$ from both sides.
$$x+15-2y=0$$
Subtract $15$ from both sides. Anything subtracted from zero gives its negation.
$$x-2y=-15$$
Consider the second equation. Subtract $y$ from both sides.
$$x-y=10$$
Put the equations in standard form and then use matrices to solve the system of equations.
For the $2\times 2$ matrix $\left(\begin{matrix}a&b\\c&d\end{matrix}\right)$, the inverse matrix is $\left(\begin{matrix}\frac{d}{ad-bc}&\frac{-b}{ad-bc}\\\frac{-c}{ad-bc}&\frac{a}{ad-bc}\end{matrix}\right)$, so the matrix equation can be rewritten as a matrix multiplication problem.
Consider the first equation. Subtract $2y$ from both sides.
$$x+15-2y=0$$
Subtract $15$ from both sides. Anything subtracted from zero gives its negation.
$$x-2y=-15$$
Consider the second equation. Subtract $y$ from both sides.
$$x-y=10$$
In order to solve by elimination, coefficients of one of the variables must be the same in both equations so that the variable will cancel out when one equation is subtracted from the other.
$$x-2y=-15,x-y=10$$
Subtract $x-y=10$ from $x-2y=-15$ by subtracting like terms on each side of the equal sign.
$$x-x-2y+y=-15-10$$
Add $x$ to $-x$. Terms $x$ and $-x$ cancel out, leaving an equation with only one variable that can be solved.
$$-2y+y=-15-10$$
Add $-2y$ to $y$.
$$-y=-15-10$$
Add $-15$ to $-10$.
$$-y=-25$$
Divide both sides by $-1$.
$$y=25$$
Substitute $25$ for $y$ in $x-y=10$. Because the resulting equation contains only one variable, you can solve for $x$ directly.
