Solve {l}{x-5=3(y-5)}{x+5=2(y+5)}

$$\left. \begin{array} { l } { x - 5 = 3 ( y - 5 ) } \\ { x + 5 = 2 ( y + 5 ) } \end{array} \right.$$

Solve for x, y

$$x=35$$

$$y=15$$

Steps Using Substitution

Consider the first equation. Use the distributive property to multiply 3 by y-5.

$$x-5=3y-15$$

Subtract 3y from both sides.

$$x-5-3y=-15$$

Add 5 to both sides.

$$x-3y=-15+5$$

Add -15 and 5 to get -10.

$$x-3y=-10$$

Consider the second equation. Use the distributive property to multiply 2 by y+5.

$$x+5=2y+10$$

Subtract 2y from both sides.

$$x+5-2y=10$$

Subtract 5 from both sides.

$$x-2y=10-5$$

Subtract 5 from 10 to get 5.

$$x-2y=5$$

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.

$$x-3y=-10,x-2y=5$$

Choose one of the equations and solve it for x by isolating x on the left hand side of the equal sign.

$$x-3y=-10$$

Add 3y to both sides of the equation.

$$x=3y-10$$

Substitute 3y-10 for x in the other equation, x-2y=5.

$$3y-10-2y=5$$

Add 3y to -2y.

$$y-10=5$$

Add 10 to both sides of the equation.

$$y=15$$

Substitute 15 for y in x=3y-10. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=3\times 15-10$$

Multiply 3 times 15.

$$x=45-10$$

Add -10 to 45.

$$x=35$$

The system is now solved.

$$x=35,y=15$$

Steps Using Matrices

Consider the first equation. Use the distributive property to multiply 3 by y-5.

$$x-5=3y-15$$

Subtract 3y from both sides.

$$x-5-3y=-15$$

Add 5 to both sides.

$$x-3y=-15+5$$

Add -15 and 5 to get -10.

$$x-3y=-10$$

Consider the second equation. Use the distributive property to multiply 2 by y+5.

$$x+5=2y+10$$

Subtract 2y from both sides.

$$x+5-2y=10$$

Subtract 5 from both sides.

$$x-2y=10-5$$

Subtract 5 from 10 to get 5.

$$x-2y=5$$

Put the equations in standard form and then use matrices to solve the system of equations.

$$x-3y=-10,x-2y=5$$

Write the equations in matrix form.

$$\left(\begin{matrix}1&-3\\1&-2\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-10\\5\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}1&-3\\1&-2\end{matrix}\right).

$$inverse(\left(\begin{matrix}1&-3\\1&-2\end{matrix}\right))\left(\begin{matrix}1&-3\\1&-2\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&-3\\1&-2\end{matrix}\right))\left(\begin{matrix}-10\\5\end{matrix}\right)$$

The product of a matrix and its inverse is the identity matrix.

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&-3\\1&-2\end{matrix}\right))\left(\begin{matrix}-10\\5\end{matrix}\right)$$

Multiply the matrices on the left hand side of the equal sign.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&-3\\1&-2\end{matrix}\right))\left(\begin{matrix}-10\\5\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{2}{-2-\left(-3\right)}&-\frac{-3}{-2-\left(-3\right)}\\-\frac{1}{-2-\left(-3\right)}&\frac{1}{-2-\left(-3\right)}\end{matrix}\right)\left(\begin{matrix}-10\\5\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-2&3\\-1&1\end{matrix}\right)\left(\begin{matrix}-10\\5\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-2\left(-10\right)+3\times 5\\-\left(-10\right)+5\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}35\\15\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=35,y=15$$

Steps Using Elimination

Consider the first equation. Use the distributive property to multiply 3 by y-5.

$$x-5=3y-15$$

Subtract 3y from both sides.

$$x-5-3y=-15$$

Add 5 to both sides.

$$x-3y=-15+5$$

Add -15 and 5 to get -10.

$$x-3y=-10$$

Consider the second equation. Use the distributive property to multiply 2 by y+5.

$$x+5=2y+10$$

Subtract 2y from both sides.

$$x+5-2y=10$$

Subtract 5 from both sides.

$$x-2y=10-5$$

Subtract 5 from 10 to get 5.

$$x-2y=5$$

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.

$$x-3y=-10,x-2y=5$$

Subtract x-2y=5 from x-3y=-10 by subtracting like terms on each side of the equal sign.

$$x-x-3y+2y=-10-5$$

Add x to -x. Terms x and -x cancel out, leaving an equation with only one variable that can be solved.

$$-3y+2y=-10-5$$

Add -3y to 2y.

$$-y=-10-5$$

Add -10 to -5.

$$-y=-15$$

Divide both sides by -1.

$$y=15$$

Substitute 15 for y in x-2y=5. Because the resulting equation contains only one variable, you can solve for x directly.

$$x-2\times 15=5$$

Multiply -2 times 15.

$$x-30=5$$

Add 30 to both sides of the equation.

$$x=35$$

The system is now solved.

$$x=35,y=15$$