Solve {l}{10-3(x+4y)=4(x+5y)}{3(x+y)=10+x+y}

$$\left\{ \begin{array} { l } { 10 - 3 ( x + 4 y ) = 4 ( x + 5 y ) } \\ { 3 ( x + y ) = 10 + x + y } \end{array} \right.$$

Solve for x, y

$$x=6$$

$$y=-1$$

Steps Using Substitution

Consider the first equation. Use the distributive property to multiply -3 by x+4y.

$$10-3x-12y=4\left(x+5y\right)$$

Use the distributive property to multiply 4 by x+5y.

$$10-3x-12y=4x+20y$$

Subtract 4x from both sides.

$$10-3x-12y-4x=20y$$

Combine -3x and -4x to get -7x.

$$10-7x-12y=20y$$

Subtract 20y from both sides.

$$10-7x-12y-20y=0$$

Combine -12y and -20y to get -32y.

$$10-7x-32y=0$$

Subtract 10 from both sides. Anything subtracted from zero gives its negation.

$$-7x-32y=-10$$

Consider the second equation. Use the distributive property to multiply 3 by x+y.

$$3x+3y=10+x+y$$

Subtract x from both sides.

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

Combine 3x and -x to get 2x.

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

Subtract y from both sides.

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

Combine 3y and -y to get 2y.

$$2x+2y=10$$

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.

$$-7x-32y=-10,2x+2y=10$$

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

$$-7x-32y=-10$$

Add 32y to both sides of the equation.

$$-7x=32y-10$$

Divide both sides by -7.

$$x=-\frac{1}{7}\left(32y-10\right)$$

Multiply -\frac{1}{7} times 32y-10.

$$x=-\frac{32}{7}y+\frac{10}{7}$$

Substitute \frac{-32y+10}{7} for x in the other equation, 2x+2y=10.

$$2\left(-\frac{32}{7}y+\frac{10}{7}\right)+2y=10$$

Multiply 2 times \frac{-32y+10}{7}.

$$-\frac{64}{7}y+\frac{20}{7}+2y=10$$

Add -\frac{64y}{7} to 2y.

$$-\frac{50}{7}y+\frac{20}{7}=10$$

Subtract \frac{20}{7} from both sides of the equation.

$$-\frac{50}{7}y=\frac{50}{7}$$

Divide both sides of the equation by -\frac{50}{7}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$y=-1$$

Substitute -1 for y in x=-\frac{32}{7}y+\frac{10}{7}. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=-\frac{32}{7}\left(-1\right)+\frac{10}{7}$$

Multiply -\frac{32}{7} times -1.

$$x=\frac{32+10}{7}$$

Add \frac{10}{7} to \frac{32}{7} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=6$$

The system is now solved.

$$x=6,y=-1$$

Steps Using Matrices

Consider the first equation. Use the distributive property to multiply -3 by x+4y.

$$10-3x-12y=4\left(x+5y\right)$$

Use the distributive property to multiply 4 by x+5y.

$$10-3x-12y=4x+20y$$

Subtract 4x from both sides.

$$10-3x-12y-4x=20y$$

Combine -3x and -4x to get -7x.

$$10-7x-12y=20y$$

Subtract 20y from both sides.

$$10-7x-12y-20y=0$$

Combine -12y and -20y to get -32y.

$$10-7x-32y=0$$

Subtract 10 from both sides. Anything subtracted from zero gives its negation.

$$-7x-32y=-10$$

Consider the second equation. Use the distributive property to multiply 3 by x+y.

$$3x+3y=10+x+y$$

Subtract x from both sides.

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

Combine 3x and -x to get 2x.

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

Subtract y from both sides.

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

Combine 3y and -y to get 2y.

$$2x+2y=10$$

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

$$-7x-32y=-10,2x+2y=10$$

Write the equations in matrix form.

$$\left(\begin{matrix}-7&-32\\2&2\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-10\\10\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}-7&-32\\2&2\end{matrix}\right).

$$inverse(\left(\begin{matrix}-7&-32\\2&2\end{matrix}\right))\left(\begin{matrix}-7&-32\\2&2\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}-7&-32\\2&2\end{matrix}\right))\left(\begin{matrix}-10\\10\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}-7&-32\\2&2\end{matrix}\right))\left(\begin{matrix}-10\\10\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}-7&-32\\2&2\end{matrix}\right))\left(\begin{matrix}-10\\10\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{2}{-7\times 2-\left(-32\times 2\right)}&-\frac{-32}{-7\times 2-\left(-32\times 2\right)}\\-\frac{2}{-7\times 2-\left(-32\times 2\right)}&-\frac{7}{-7\times 2-\left(-32\times 2\right)}\end{matrix}\right)\left(\begin{matrix}-10\\10\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{25}&\frac{16}{25}\\-\frac{1}{25}&-\frac{7}{50}\end{matrix}\right)\left(\begin{matrix}-10\\10\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{25}\left(-10\right)+\frac{16}{25}\times 10\\-\frac{1}{25}\left(-10\right)-\frac{7}{50}\times 10\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}6\\-1\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=6,y=-1$$

Steps Using Elimination

Consider the first equation. Use the distributive property to multiply -3 by x+4y.

$$10-3x-12y=4\left(x+5y\right)$$

Use the distributive property to multiply 4 by x+5y.

$$10-3x-12y=4x+20y$$

Subtract 4x from both sides.

$$10-3x-12y-4x=20y$$

Combine -3x and -4x to get -7x.

$$10-7x-12y=20y$$

Subtract 20y from both sides.

$$10-7x-12y-20y=0$$

Combine -12y and -20y to get -32y.

$$10-7x-32y=0$$

Subtract 10 from both sides. Anything subtracted from zero gives its negation.

$$-7x-32y=-10$$

Consider the second equation. Use the distributive property to multiply 3 by x+y.

$$3x+3y=10+x+y$$

Subtract x from both sides.

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

Combine 3x and -x to get 2x.

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

Subtract y from both sides.

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

Combine 3y and -y to get 2y.

$$2x+2y=10$$

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.

$$-7x-32y=-10,2x+2y=10$$

To make -7x and 2x equal, multiply all terms on each side of the first equation by 2 and all terms on each side of the second by -7.

$$2\left(-7\right)x+2\left(-32\right)y=2\left(-10\right),-7\times 2x-7\times 2y=-7\times 10$$

Simplify.

$$-14x-64y=-20,-14x-14y=-70$$

Subtract -14x-14y=-70 from -14x-64y=-20 by subtracting like terms on each side of the equal sign.

$$-14x+14x-64y+14y=-20+70$$

Add -14x to 14x. Terms -14x and 14x cancel out, leaving an equation with only one variable that can be solved.

$$-64y+14y=-20+70$$

Add -64y to 14y.

$$-50y=-20+70$$

Add -20 to 70.

$$-50y=50$$

Divide both sides by -50.

$$y=-1$$

Substitute -1 for y in 2x+2y=10. Because the resulting equation contains only one variable, you can solve for x directly.

$$2x+2\left(-1\right)=10$$

Multiply 2 times -1.

$$2x-2=10$$

Add 2 to both sides of the equation.

$$2x=12$$

Divide both sides by 2.

$$x=6$$

The system is now solved.

$$x=6,y=-1$$