Solve {l}{4(x+y)-3(x-y)=10}{2(x+y)-3(x-y)=2}

$$\left\{ \begin{array} { l } { 4 ( x + y ) - 3 ( x - y ) = 10 } \\ { 2 ( x + y ) - 3 ( x - y ) = 2 } \end{array} \right.$$

Solve for x, y

$$x=3$$

$$y=1$$

Steps Using Substitution

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

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

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

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

Combine 4x and -3x to get x.

$$x+4y+3y=10$$

Combine 4y and 3y to get 7y.

$$x+7y=10$$

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

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

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

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

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

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

Combine 2y and 3y to get 5y.

$$-x+5y=2$$

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.

$$x+7y=10,-x+5y=2$$

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

$$x+7y=10$$

Subtract 7y from both sides of the equation.

$$x=-7y+10$$

Substitute -7y+10 for x in the other equation, -x+5y=2.

$$-\left(-7y+10\right)+5y=2$$

Multiply -1 times -7y+10.

$$7y-10+5y=2$$

Add 7y to 5y.

$$12y-10=2$$

Add 10 to both sides of the equation.

$$12y=12$$

Divide both sides by 12.

$$y=1$$

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

$$x=-7+10$$

Add 10 to -7.

$$x=3$$

The system is now solved.

$$x=3,y=1$$

Steps Using Matrices

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

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

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

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

Combine 4x and -3x to get x.

$$x+4y+3y=10$$

Combine 4y and 3y to get 7y.

$$x+7y=10$$

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

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

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

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

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

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

Combine 2y and 3y to get 5y.

$$-x+5y=2$$

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

$$x+7y=10,-x+5y=2$$

Write the equations in matrix form.

$$\left(\begin{matrix}1&7\\-1&5\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}10\\2\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}1&7\\-1&5\end{matrix}\right).

$$inverse(\left(\begin{matrix}1&7\\-1&5\end{matrix}\right))\left(\begin{matrix}1&7\\-1&5\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&7\\-1&5\end{matrix}\right))\left(\begin{matrix}10\\2\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&7\\-1&5\end{matrix}\right))\left(\begin{matrix}10\\2\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&7\\-1&5\end{matrix}\right))\left(\begin{matrix}10\\2\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{5}{5-7\left(-1\right)}&-\frac{7}{5-7\left(-1\right)}\\-\frac{-1}{5-7\left(-1\right)}&\frac{1}{5-7\left(-1\right)}\end{matrix}\right)\left(\begin{matrix}10\\2\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{5}{12}&-\frac{7}{12}\\\frac{1}{12}&\frac{1}{12}\end{matrix}\right)\left(\begin{matrix}10\\2\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{5}{12}\times 10-\frac{7}{12}\times 2\\\frac{1}{12}\times 10+\frac{1}{12}\times 2\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}3\\1\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=3,y=1$$

Steps Using Elimination

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

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

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

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

Combine 4x and -3x to get x.

$$x+4y+3y=10$$

Combine 4y and 3y to get 7y.

$$x+7y=10$$

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

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

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

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

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

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

Combine 2y and 3y to get 5y.

$$-x+5y=2$$

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.

$$x+7y=10,-x+5y=2$$

To make x and -x equal, multiply all terms on each side of the first equation by -1 and all terms on each side of the second by 1.

$$-x-7y=-10,-x+5y=2$$

Subtract -x+5y=2 from -x-7y=-10 by subtracting like terms on each side of the equal sign.

$$-x+x-7y-5y=-10-2$$

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

$$-7y-5y=-10-2$$

Add -7y to -5y.

$$-12y=-10-2$$

Add -10 to -2.

$$-12y=-12$$

Divide both sides by -12.

$$y=1$$

Substitute 1 for y in -x+5y=2. Because the resulting equation contains only one variable, you can solve for x directly.

$$-x+5=2$$

Subtract 5 from both sides of the equation.

$$-x=-3$$

Divide both sides by -1.

$$x=3$$

The system is now solved.

$$x=3,y=1$$