Solve {r}{2(x+y)=61-2y}{4x-4=18-3y}

$$\left. \begin{array} { r } { 2 ( x + y ) = 61 - 2 y } \\ { 4 x - 4 = 18 - 3 y } \end{array} \right.$$

Solve for x, y

$$x = -\frac{19}{2} = -9\frac{1}{2} = -9.5$$

$$y=20$$

Steps Using Substitution

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

$$2x+2y=61-2y$$

Add 2y to both sides.

$$2x+2y+2y=61$$

Combine 2y and 2y to get 4y.

$$2x+4y=61$$

Consider the second equation. Add 3y to both sides.

$$4x-4+3y=18$$

Add 4 to both sides.

$$4x+3y=18+4$$

Add 18 and 4 to get 22.

$$4x+3y=22$$

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.

$$2x+4y=61,4x+3y=22$$

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

$$2x+4y=61$$

Subtract 4y from both sides of the equation.

$$2x=-4y+61$$

Divide both sides by 2.

$$x=\frac{1}{2}\left(-4y+61\right)$$

Multiply \frac{1}{2} times -4y+61.

$$x=-2y+\frac{61}{2}$$

Substitute -2y+\frac{61}{2} for x in the other equation, 4x+3y=22.

$$4\left(-2y+\frac{61}{2}\right)+3y=22$$

Multiply 4 times -2y+\frac{61}{2}.

$$-8y+122+3y=22$$

Add -8y to 3y.

$$-5y+122=22$$

Subtract 122 from both sides of the equation.

$$-5y=-100$$

Divide both sides by -5.

$$y=20$$

Substitute 20 for y in x=-2y+\frac{61}{2}. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=-2\times 20+\frac{61}{2}$$

Multiply -2 times 20.

$$x=-40+\frac{61}{2}$$

Add \frac{61}{2} to -40.

$$x=-\frac{19}{2}$$

The system is now solved.

$$x=-\frac{19}{2},y=20$$

Steps Using Matrices

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

$$2x+2y=61-2y$$

Add 2y to both sides.

$$2x+2y+2y=61$$

Combine 2y and 2y to get 4y.

$$2x+4y=61$$

Consider the second equation. Add 3y to both sides.

$$4x-4+3y=18$$

Add 4 to both sides.

$$4x+3y=18+4$$

Add 18 and 4 to get 22.

$$4x+3y=22$$

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

$$2x+4y=61,4x+3y=22$$

Write the equations in matrix form.

$$\left(\begin{matrix}2&4\\4&3\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}61\\22\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}2&4\\4&3\end{matrix}\right).

$$inverse(\left(\begin{matrix}2&4\\4&3\end{matrix}\right))\left(\begin{matrix}2&4\\4&3\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}2&4\\4&3\end{matrix}\right))\left(\begin{matrix}61\\22\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}2&4\\4&3\end{matrix}\right))\left(\begin{matrix}61\\22\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}2&4\\4&3\end{matrix}\right))\left(\begin{matrix}61\\22\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{3}{2\times 3-4\times 4}&-\frac{4}{2\times 3-4\times 4}\\-\frac{4}{2\times 3-4\times 4}&\frac{2}{2\times 3-4\times 4}\end{matrix}\right)\left(\begin{matrix}61\\22\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{3}{10}&\frac{2}{5}\\\frac{2}{5}&-\frac{1}{5}\end{matrix}\right)\left(\begin{matrix}61\\22\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{3}{10}\times 61+\frac{2}{5}\times 22\\\frac{2}{5}\times 61-\frac{1}{5}\times 22\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{19}{2}\\20\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=-\frac{19}{2},y=20$$

Steps Using Elimination

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

$$2x+2y=61-2y$$

Add 2y to both sides.

$$2x+2y+2y=61$$

Combine 2y and 2y to get 4y.

$$2x+4y=61$$

Consider the second equation. Add 3y to both sides.

$$4x-4+3y=18$$

Add 4 to both sides.

$$4x+3y=18+4$$

Add 18 and 4 to get 22.

$$4x+3y=22$$

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.

$$2x+4y=61,4x+3y=22$$

To make 2x and 4x equal, multiply all terms on each side of the first equation by 4 and all terms on each side of the second by 2.

$$4\times 2x+4\times 4y=4\times 61,2\times 4x+2\times 3y=2\times 22$$

Simplify.

$$8x+16y=244,8x+6y=44$$

Subtract 8x+6y=44 from 8x+16y=244 by subtracting like terms on each side of the equal sign.

$$8x-8x+16y-6y=244-44$$

Add 8x to -8x. Terms 8x and -8x cancel out, leaving an equation with only one variable that can be solved.

$$16y-6y=244-44$$

Add 16y to -6y.

$$10y=244-44$$

Add 244 to -44.

$$10y=200$$

Divide both sides by 10.

$$y=20$$

Substitute 20 for y in 4x+3y=22. Because the resulting equation contains only one variable, you can solve for x directly.

$$4x+3\times 20=22$$

Multiply 3 times 20.

$$4x+60=22$$

Subtract 60 from both sides of the equation.

$$4x=-38$$

Divide both sides by 4.

$$x=-\frac{19}{2}$$

The system is now solved.

$$x=-\frac{19}{2},y=20$$