Solve {l}{3(x+y)-5(x-y)=16}{2(x+y)+x-y=15}

$$\left\{ \begin{array} { l } { 3 ( x + y ) - 5 ( x - y ) = 16 } \\ { 2 ( x + y ) + x - y = 15 } \end{array} \right.$$

Solve for x, y

$$x=4$$

$$y=3$$

Steps Using Substitution

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

$$3x+3y-5\left(x-y\right)=16$$

Use the distributive property to multiply -5 by x-y.

$$3x+3y-5x+5y=16$$

Combine 3x and -5x to get -2x.

$$-2x+3y+5y=16$$

Combine 3y and 5y to get 8y.

$$-2x+8y=16$$

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

$$2x+2y+x-y=15$$

Combine 2x and x to get 3x.

$$3x+2y-y=15$$

Combine 2y and -y to get y.

$$3x+y=15$$

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.

$$-2x+8y=16,3x+y=15$$

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

$$-2x+8y=16$$

Subtract 8y from both sides of the equation.

$$-2x=-8y+16$$

Divide both sides by -2.

$$x=-\frac{1}{2}\left(-8y+16\right)$$

Multiply -\frac{1}{2} times -8y+16.

$$x=4y-8$$

Substitute -8+4y for x in the other equation, 3x+y=15.

$$3\left(4y-8\right)+y=15$$

Multiply 3 times -8+4y.

$$12y-24+y=15$$

Add 12y to y.

$$13y-24=15$$

Add 24 to both sides of the equation.

$$13y=39$$

Divide both sides by 13.

$$y=3$$

Substitute 3 for y in x=4y-8. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=4\times 3-8$$

Multiply 4 times 3.

$$x=12-8$$

Add -8 to 12.

$$x=4$$

The system is now solved.

$$x=4,y=3$$

Steps Using Matrices

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

$$3x+3y-5\left(x-y\right)=16$$

Use the distributive property to multiply -5 by x-y.

$$3x+3y-5x+5y=16$$

Combine 3x and -5x to get -2x.

$$-2x+3y+5y=16$$

Combine 3y and 5y to get 8y.

$$-2x+8y=16$$

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

$$2x+2y+x-y=15$$

Combine 2x and x to get 3x.

$$3x+2y-y=15$$

Combine 2y and -y to get y.

$$3x+y=15$$

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

$$-2x+8y=16,3x+y=15$$

Write the equations in matrix form.

$$\left(\begin{matrix}-2&8\\3&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}16\\15\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}-2&8\\3&1\end{matrix}\right).

$$inverse(\left(\begin{matrix}-2&8\\3&1\end{matrix}\right))\left(\begin{matrix}-2&8\\3&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}-2&8\\3&1\end{matrix}\right))\left(\begin{matrix}16\\15\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}-2&8\\3&1\end{matrix}\right))\left(\begin{matrix}16\\15\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}-2&8\\3&1\end{matrix}\right))\left(\begin{matrix}16\\15\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{-2-8\times 3}&-\frac{8}{-2-8\times 3}\\-\frac{3}{-2-8\times 3}&-\frac{2}{-2-8\times 3}\end{matrix}\right)\left(\begin{matrix}16\\15\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{26}&\frac{4}{13}\\\frac{3}{26}&\frac{1}{13}\end{matrix}\right)\left(\begin{matrix}16\\15\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{26}\times 16+\frac{4}{13}\times 15\\\frac{3}{26}\times 16+\frac{1}{13}\times 15\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}4\\3\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=4,y=3$$

Steps Using Elimination

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

$$3x+3y-5\left(x-y\right)=16$$

Use the distributive property to multiply -5 by x-y.

$$3x+3y-5x+5y=16$$

Combine 3x and -5x to get -2x.

$$-2x+3y+5y=16$$

Combine 3y and 5y to get 8y.

$$-2x+8y=16$$

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

$$2x+2y+x-y=15$$

Combine 2x and x to get 3x.

$$3x+2y-y=15$$

Combine 2y and -y to get y.

$$3x+y=15$$

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.

$$-2x+8y=16,3x+y=15$$

To make -2x and 3x equal, multiply all terms on each side of the first equation by 3 and all terms on each side of the second by -2.

$$3\left(-2\right)x+3\times 8y=3\times 16,-2\times 3x-2y=-2\times 15$$

Simplify.

$$-6x+24y=48,-6x-2y=-30$$

Subtract -6x-2y=-30 from -6x+24y=48 by subtracting like terms on each side of the equal sign.

$$-6x+6x+24y+2y=48+30$$

Add -6x to 6x. Terms -6x and 6x cancel out, leaving an equation with only one variable that can be solved.

$$24y+2y=48+30$$

Add 24y to 2y.

$$26y=48+30$$

Add 48 to 30.

$$26y=78$$

Divide both sides by 26.

$$y=3$$

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

$$3x+3=15$$

Subtract 3 from both sides of the equation.

$$3x=12$$

Divide both sides by 3.

$$x=4$$

The system is now solved.

$$x=4,y=3$$