Solve {l}{7(y-4x)+2(x-11)=2(x-18)}{3(20-y)+15x=6x+27}

$$\left\{ \begin{array} { l } { 7 ( y - 4 x ) + 2 ( x - 11 ) = 2 ( x - 18 ) } \\ { 3 ( 20 - y ) + 15 x = 6 x + 27 } \end{array} \right.$$

Solve for y, x

$$x=13$$

$$y=50$$

Steps Using Substitution

Consider the first equation. Use the distributive property to multiply 7 by y-4x.

$$7y-28x+2\left(x-11\right)=2\left(x-18\right)$$

Use the distributive property to multiply 2 by x-11.

$$7y-28x+2x-22=2\left(x-18\right)$$

Combine -28x and 2x to get -26x.

$$7y-26x-22=2\left(x-18\right)$$

Use the distributive property to multiply 2 by x-18.

$$7y-26x-22=2x-36$$

Subtract 2x from both sides.

$$7y-26x-22-2x=-36$$

Combine -26x and -2x to get -28x.

$$7y-28x-22=-36$$

Add 22 to both sides.

$$7y-28x=-36+22$$

Add -36 and 22 to get -14.

$$7y-28x=-14$$

Consider the second equation. Use the distributive property to multiply 3 by 20-y.

$$60-3y+15x=6x+27$$

Subtract 6x from both sides.

$$60-3y+15x-6x=27$$

Combine 15x and -6x to get 9x.

$$60-3y+9x=27$$

Subtract 60 from both sides.

$$-3y+9x=27-60$$

Subtract 60 from 27 to get -33.

$$-3y+9x=-33$$

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.

$$7y-28x=-14,-3y+9x=-33$$

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

$$7y-28x=-14$$

Add 28x to both sides of the equation.

$$7y=28x-14$$

Divide both sides by 7.

$$y=\frac{1}{7}\left(28x-14\right)$$

Multiply \frac{1}{7} times 28x-14.

$$y=4x-2$$

Substitute 4x-2 for y in the other equation, -3y+9x=-33.

$$-3\left(4x-2\right)+9x=-33$$

Multiply -3 times 4x-2.

$$-12x+6+9x=-33$$

Add -12x to 9x.

$$-3x+6=-33$$

Subtract 6 from both sides of the equation.

$$-3x=-39$$

Divide both sides by -3.

$$x=13$$

Substitute 13 for x in y=4x-2. Because the resulting equation contains only one variable, you can solve for y directly.

$$y=4\times 13-2$$

Multiply 4 times 13.

$$y=52-2$$

Add -2 to 52.

$$y=50$$

The system is now solved.

$$y=50,x=13$$

Steps Using Matrices

Consider the first equation. Use the distributive property to multiply 7 by y-4x.

$$7y-28x+2\left(x-11\right)=2\left(x-18\right)$$

Use the distributive property to multiply 2 by x-11.

$$7y-28x+2x-22=2\left(x-18\right)$$

Combine -28x and 2x to get -26x.

$$7y-26x-22=2\left(x-18\right)$$

Use the distributive property to multiply 2 by x-18.

$$7y-26x-22=2x-36$$

Subtract 2x from both sides.

$$7y-26x-22-2x=-36$$

Combine -26x and -2x to get -28x.

$$7y-28x-22=-36$$

Add 22 to both sides.

$$7y-28x=-36+22$$

Add -36 and 22 to get -14.

$$7y-28x=-14$$

Consider the second equation. Use the distributive property to multiply 3 by 20-y.

$$60-3y+15x=6x+27$$

Subtract 6x from both sides.

$$60-3y+15x-6x=27$$

Combine 15x and -6x to get 9x.

$$60-3y+9x=27$$

Subtract 60 from both sides.

$$-3y+9x=27-60$$

Subtract 60 from 27 to get -33.

$$-3y+9x=-33$$

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

$$7y-28x=-14,-3y+9x=-33$$

Write the equations in matrix form.

$$\left(\begin{matrix}7&-28\\-3&9\end{matrix}\right)\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}-14\\-33\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}7&-28\\-3&9\end{matrix}\right).

$$inverse(\left(\begin{matrix}7&-28\\-3&9\end{matrix}\right))\left(\begin{matrix}7&-28\\-3&9\end{matrix}\right)\left(\begin{matrix}y\\x\end{matrix}\right)=inverse(\left(\begin{matrix}7&-28\\-3&9\end{matrix}\right))\left(\begin{matrix}-14\\-33\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}y\\x\end{matrix}\right)=inverse(\left(\begin{matrix}7&-28\\-3&9\end{matrix}\right))\left(\begin{matrix}-14\\-33\end{matrix}\right)$$

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

$$\left(\begin{matrix}y\\x\end{matrix}\right)=inverse(\left(\begin{matrix}7&-28\\-3&9\end{matrix}\right))\left(\begin{matrix}-14\\-33\end{matrix}\right)$$

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.

$$\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}\frac{9}{7\times 9-\left(-28\left(-3\right)\right)}&-\frac{-28}{7\times 9-\left(-28\left(-3\right)\right)}\\-\frac{-3}{7\times 9-\left(-28\left(-3\right)\right)}&\frac{7}{7\times 9-\left(-28\left(-3\right)\right)}\end{matrix}\right)\left(\begin{matrix}-14\\-33\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}-\frac{3}{7}&-\frac{4}{3}\\-\frac{1}{7}&-\frac{1}{3}\end{matrix}\right)\left(\begin{matrix}-14\\-33\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}-\frac{3}{7}\left(-14\right)-\frac{4}{3}\left(-33\right)\\-\frac{1}{7}\left(-14\right)-\frac{1}{3}\left(-33\right)\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}50\\13\end{matrix}\right)$$

Extract the matrix elements y and x.

$$y=50,x=13$$

Steps Using Elimination

Consider the first equation. Use the distributive property to multiply 7 by y-4x.

$$7y-28x+2\left(x-11\right)=2\left(x-18\right)$$

Use the distributive property to multiply 2 by x-11.

$$7y-28x+2x-22=2\left(x-18\right)$$

Combine -28x and 2x to get -26x.

$$7y-26x-22=2\left(x-18\right)$$

Use the distributive property to multiply 2 by x-18.

$$7y-26x-22=2x-36$$

Subtract 2x from both sides.

$$7y-26x-22-2x=-36$$

Combine -26x and -2x to get -28x.

$$7y-28x-22=-36$$

Add 22 to both sides.

$$7y-28x=-36+22$$

Add -36 and 22 to get -14.

$$7y-28x=-14$$

Consider the second equation. Use the distributive property to multiply 3 by 20-y.

$$60-3y+15x=6x+27$$

Subtract 6x from both sides.

$$60-3y+15x-6x=27$$

Combine 15x and -6x to get 9x.

$$60-3y+9x=27$$

Subtract 60 from both sides.

$$-3y+9x=27-60$$

Subtract 60 from 27 to get -33.

$$-3y+9x=-33$$

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.

$$7y-28x=-14,-3y+9x=-33$$

To make 7y and -3y equal, multiply all terms on each side of the first equation by -3 and all terms on each side of the second by 7.

$$-3\times 7y-3\left(-28\right)x=-3\left(-14\right),7\left(-3\right)y+7\times 9x=7\left(-33\right)$$

Simplify.

$$-21y+84x=42,-21y+63x=-231$$

Subtract -21y+63x=-231 from -21y+84x=42 by subtracting like terms on each side of the equal sign.

$$-21y+21y+84x-63x=42+231$$

Add -21y to 21y. Terms -21y and 21y cancel out, leaving an equation with only one variable that can be solved.

$$84x-63x=42+231$$

Add 84x to -63x.

$$21x=42+231$$

Add 42 to 231.

$$21x=273$$

Divide both sides by 21.

$$x=13$$

Substitute 13 for x in -3y+9x=-33. Because the resulting equation contains only one variable, you can solve for y directly.

$$-3y+9\times 13=-33$$

Multiply 9 times 13.

$$-3y+117=-33$$

Subtract 117 from both sides of the equation.

$$-3y=-150$$

Divide both sides by -3.

$$y=50$$

The system is now solved.

$$y=50,x=13$$