Solve {l}{2(4x+21)+x=y-3(2x-y)}{2(3x-1)-7y=-3x+4(y+2)-61}

$$\left\{ \begin{array} { l } { 2 ( 4 x + 21 ) + x = y - 3 ( 2 x - y ) } \\ { 2 ( 3 x - 1 ) - 7 y = - 3 x + 4 ( y + 2 ) - 61 } \end{array} \right.$$

Solve for x, y

$$x=-2$$

$$y=3$$

Steps Using Substitution

Consider the first equation. Use the distributive property to multiply 2 by 4x+21.

$$8x+42+x=y-3\left(2x-y\right)$$

Combine 8x and x to get 9x.

$$9x+42=y-3\left(2x-y\right)$$

Use the distributive property to multiply -3 by 2x-y.

$$9x+42=y-6x+3y$$

Combine y and 3y to get 4y.

$$9x+42=4y-6x$$

Subtract 4y from both sides.

$$9x+42-4y=-6x$$

Add 6x to both sides.

$$9x+42-4y+6x=0$$

Combine 9x and 6x to get 15x.

$$15x+42-4y=0$$

Subtract 42 from both sides. Anything subtracted from zero gives its negation.

$$15x-4y=-42$$

Consider the second equation. Use the distributive property to multiply 2 by 3x-1.

$$6x-2-7y=-3x+4\left(y+2\right)-61$$

Use the distributive property to multiply 4 by y+2.

$$6x-2-7y=-3x+4y+8-61$$

Subtract 61 from 8 to get -53.

$$6x-2-7y=-3x+4y-53$$

Add 3x to both sides.

$$6x-2-7y+3x=4y-53$$

Combine 6x and 3x to get 9x.

$$9x-2-7y=4y-53$$

Subtract 4y from both sides.

$$9x-2-7y-4y=-53$$

Combine -7y and -4y to get -11y.

$$9x-2-11y=-53$$

Add 2 to both sides.

$$9x-11y=-53+2$$

Add -53 and 2 to get -51.

$$9x-11y=-51$$

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.

$$15x-4y=-42,9x-11y=-51$$

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

$$15x-4y=-42$$

Add 4y to both sides of the equation.

$$15x=4y-42$$

Divide both sides by 15.

$$x=\frac{1}{15}\left(4y-42\right)$$

Multiply \frac{1}{15} times 4y-42.

$$x=\frac{4}{15}y-\frac{14}{5}$$

Substitute \frac{4y}{15}-\frac{14}{5} for x in the other equation, 9x-11y=-51.

$$9\left(\frac{4}{15}y-\frac{14}{5}\right)-11y=-51$$

Multiply 9 times \frac{4y}{15}-\frac{14}{5}.

$$\frac{12}{5}y-\frac{126}{5}-11y=-51$$

Add \frac{12y}{5} to -11y.

$$-\frac{43}{5}y-\frac{126}{5}=-51$$

Add \frac{126}{5} to both sides of the equation.

$$-\frac{43}{5}y=-\frac{129}{5}$$

Divide both sides of the equation by -\frac{43}{5}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$y=3$$

Substitute 3 for y in x=\frac{4}{15}y-\frac{14}{5}. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=\frac{4}{15}\times 3-\frac{14}{5}$$

Multiply \frac{4}{15} times 3.

$$x=\frac{4-14}{5}$$

Add -\frac{14}{5} to \frac{4}{5} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=-2$$

The system is now solved.

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

Steps Using Matrices

Consider the first equation. Use the distributive property to multiply 2 by 4x+21.

$$8x+42+x=y-3\left(2x-y\right)$$

Combine 8x and x to get 9x.

$$9x+42=y-3\left(2x-y\right)$$

Use the distributive property to multiply -3 by 2x-y.

$$9x+42=y-6x+3y$$

Combine y and 3y to get 4y.

$$9x+42=4y-6x$$

Subtract 4y from both sides.

$$9x+42-4y=-6x$$

Add 6x to both sides.

$$9x+42-4y+6x=0$$

Combine 9x and 6x to get 15x.

$$15x+42-4y=0$$

Subtract 42 from both sides. Anything subtracted from zero gives its negation.

$$15x-4y=-42$$

Consider the second equation. Use the distributive property to multiply 2 by 3x-1.

$$6x-2-7y=-3x+4\left(y+2\right)-61$$

Use the distributive property to multiply 4 by y+2.

$$6x-2-7y=-3x+4y+8-61$$

Subtract 61 from 8 to get -53.

$$6x-2-7y=-3x+4y-53$$

Add 3x to both sides.

$$6x-2-7y+3x=4y-53$$

Combine 6x and 3x to get 9x.

$$9x-2-7y=4y-53$$

Subtract 4y from both sides.

$$9x-2-7y-4y=-53$$

Combine -7y and -4y to get -11y.

$$9x-2-11y=-53$$

Add 2 to both sides.

$$9x-11y=-53+2$$

Add -53 and 2 to get -51.

$$9x-11y=-51$$

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

$$15x-4y=-42,9x-11y=-51$$

Write the equations in matrix form.

$$\left(\begin{matrix}15&-4\\9&-11\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-42\\-51\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}15&-4\\9&-11\end{matrix}\right).

$$inverse(\left(\begin{matrix}15&-4\\9&-11\end{matrix}\right))\left(\begin{matrix}15&-4\\9&-11\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}15&-4\\9&-11\end{matrix}\right))\left(\begin{matrix}-42\\-51\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}15&-4\\9&-11\end{matrix}\right))\left(\begin{matrix}-42\\-51\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}15&-4\\9&-11\end{matrix}\right))\left(\begin{matrix}-42\\-51\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{11}{15\left(-11\right)-\left(-4\times 9\right)}&-\frac{-4}{15\left(-11\right)-\left(-4\times 9\right)}\\-\frac{9}{15\left(-11\right)-\left(-4\times 9\right)}&\frac{15}{15\left(-11\right)-\left(-4\times 9\right)}\end{matrix}\right)\left(\begin{matrix}-42\\-51\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{11}{129}&-\frac{4}{129}\\\frac{3}{43}&-\frac{5}{43}\end{matrix}\right)\left(\begin{matrix}-42\\-51\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{11}{129}\left(-42\right)-\frac{4}{129}\left(-51\right)\\\frac{3}{43}\left(-42\right)-\frac{5}{43}\left(-51\right)\end{matrix}\right)$$

Do the arithmetic.

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

Extract the matrix elements x and y.

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

Steps Using Elimination

Consider the first equation. Use the distributive property to multiply 2 by 4x+21.

$$8x+42+x=y-3\left(2x-y\right)$$

Combine 8x and x to get 9x.

$$9x+42=y-3\left(2x-y\right)$$

Use the distributive property to multiply -3 by 2x-y.

$$9x+42=y-6x+3y$$

Combine y and 3y to get 4y.

$$9x+42=4y-6x$$

Subtract 4y from both sides.

$$9x+42-4y=-6x$$

Add 6x to both sides.

$$9x+42-4y+6x=0$$

Combine 9x and 6x to get 15x.

$$15x+42-4y=0$$

Subtract 42 from both sides. Anything subtracted from zero gives its negation.

$$15x-4y=-42$$

Consider the second equation. Use the distributive property to multiply 2 by 3x-1.

$$6x-2-7y=-3x+4\left(y+2\right)-61$$

Use the distributive property to multiply 4 by y+2.

$$6x-2-7y=-3x+4y+8-61$$

Subtract 61 from 8 to get -53.

$$6x-2-7y=-3x+4y-53$$

Add 3x to both sides.

$$6x-2-7y+3x=4y-53$$

Combine 6x and 3x to get 9x.

$$9x-2-7y=4y-53$$

Subtract 4y from both sides.

$$9x-2-7y-4y=-53$$

Combine -7y and -4y to get -11y.

$$9x-2-11y=-53$$

Add 2 to both sides.

$$9x-11y=-53+2$$

Add -53 and 2 to get -51.

$$9x-11y=-51$$

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.

$$15x-4y=-42,9x-11y=-51$$

To make 15x and 9x equal, multiply all terms on each side of the first equation by 9 and all terms on each side of the second by 15.

$$9\times 15x+9\left(-4\right)y=9\left(-42\right),15\times 9x+15\left(-11\right)y=15\left(-51\right)$$

Simplify.

$$135x-36y=-378,135x-165y=-765$$

Subtract 135x-165y=-765 from 135x-36y=-378 by subtracting like terms on each side of the equal sign.

$$135x-135x-36y+165y=-378+765$$

Add 135x to -135x. Terms 135x and -135x cancel out, leaving an equation with only one variable that can be solved.

$$-36y+165y=-378+765$$

Add -36y to 165y.

$$129y=-378+765$$

Add -378 to 765.

$$129y=387$$

Divide both sides by 129.

$$y=3$$

Substitute 3 for y in 9x-11y=-51. Because the resulting equation contains only one variable, you can solve for x directly.

$$9x-11\times 3=-51$$

Multiply -11 times 3.

$$9x-33=-51$$

Add 33 to both sides of the equation.

$$9x=-18$$

Divide both sides by 9.

$$x=-2$$

The system is now solved.

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