Solve {l}{x+5=4(y+5)}{x-2=11(y-2)}

$$\left\{ \begin{array} { l } { x + 5 = 4 ( y + 5 ) } \\ { x - 2 = 11 ( y - 2 ) } \end{array} \right.$$

Solve for x, y

$$x=35$$

$$y=5$$

Steps Using Substitution

Consider the first equation. Use the distributive property to multiply 4 by y+5.

$$x+5=4y+20$$

Subtract 4y from both sides.

$$x+5-4y=20$$

Subtract 5 from both sides.

$$x-4y=20-5$$

Subtract 5 from 20 to get 15.

$$x-4y=15$$

Consider the second equation. Use the distributive property to multiply 11 by y-2.

$$x-2=11y-22$$

Subtract 11y from both sides.

$$x-2-11y=-22$$

Add 2 to both sides.

$$x-11y=-22+2$$

Add -22 and 2 to get -20.

$$x-11y=-20$$

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.

$$x-4y=15,x-11y=-20$$

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

$$x-4y=15$$

Add 4y to both sides of the equation.

$$x=4y+15$$

Substitute 4y+15 for x in the other equation, x-11y=-20.

$$4y+15-11y=-20$$

Add 4y to -11y.

$$-7y+15=-20$$

Subtract 15 from both sides of the equation.

$$-7y=-35$$

Divide both sides by -7.

$$y=5$$

Substitute 5 for y in x=4y+15. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=4\times 5+15$$

Multiply 4 times 5.

$$x=20+15$$

Add 15 to 20.

$$x=35$$

The system is now solved.

$$x=35,y=5$$

Steps Using Matrices

Consider the first equation. Use the distributive property to multiply 4 by y+5.

$$x+5=4y+20$$

Subtract 4y from both sides.

$$x+5-4y=20$$

Subtract 5 from both sides.

$$x-4y=20-5$$

Subtract 5 from 20 to get 15.

$$x-4y=15$$

Consider the second equation. Use the distributive property to multiply 11 by y-2.

$$x-2=11y-22$$

Subtract 11y from both sides.

$$x-2-11y=-22$$

Add 2 to both sides.

$$x-11y=-22+2$$

Add -22 and 2 to get -20.

$$x-11y=-20$$

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

$$x-4y=15,x-11y=-20$$

Write the equations in matrix form.

$$\left(\begin{matrix}1&-4\\1&-11\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}15\\-20\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}1&-4\\1&-11\end{matrix}\right).

$$inverse(\left(\begin{matrix}1&-4\\1&-11\end{matrix}\right))\left(\begin{matrix}1&-4\\1&-11\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&-4\\1&-11\end{matrix}\right))\left(\begin{matrix}15\\-20\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&-4\\1&-11\end{matrix}\right))\left(\begin{matrix}15\\-20\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&-4\\1&-11\end{matrix}\right))\left(\begin{matrix}15\\-20\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{11}{-11-\left(-4\right)}&-\frac{-4}{-11-\left(-4\right)}\\-\frac{1}{-11-\left(-4\right)}&\frac{1}{-11-\left(-4\right)}\end{matrix}\right)\left(\begin{matrix}15\\-20\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{11}{7}&-\frac{4}{7}\\\frac{1}{7}&-\frac{1}{7}\end{matrix}\right)\left(\begin{matrix}15\\-20\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{11}{7}\times 15-\frac{4}{7}\left(-20\right)\\\frac{1}{7}\times 15-\frac{1}{7}\left(-20\right)\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}35\\5\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=35,y=5$$

Steps Using Elimination

Consider the first equation. Use the distributive property to multiply 4 by y+5.

$$x+5=4y+20$$

Subtract 4y from both sides.

$$x+5-4y=20$$

Subtract 5 from both sides.

$$x-4y=20-5$$

Subtract 5 from 20 to get 15.

$$x-4y=15$$

Consider the second equation. Use the distributive property to multiply 11 by y-2.

$$x-2=11y-22$$

Subtract 11y from both sides.

$$x-2-11y=-22$$

Add 2 to both sides.

$$x-11y=-22+2$$

Add -22 and 2 to get -20.

$$x-11y=-20$$

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.

$$x-4y=15,x-11y=-20$$

Subtract x-11y=-20 from x-4y=15 by subtracting like terms on each side of the equal sign.

$$x-x-4y+11y=15+20$$

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

$$-4y+11y=15+20$$

Add -4y to 11y.

$$7y=15+20$$

Add 15 to 20.

$$7y=35$$

Divide both sides by 7.

$$y=5$$

Substitute 5 for y in x-11y=-20. Because the resulting equation contains only one variable, you can solve for x directly.

$$x-11\times 5=-20$$

Multiply -11 times 5.

$$x-55=-20$$

Add 55 to both sides of the equation.

$$x=35$$

The system is now solved.

$$x=35,y=5$$