Solve {l}{x+y=360}{4(x+40)+2(1.5y-60)=1320}

$$\left\{ \begin{array} { l } { x + y = 360 } \\ { 4 ( x + 40 ) + 2 ( 1.5 y - 60 ) = 1320 } \end{array} \right.$$

Solve for x, y

$$x=200$$

$$y=160$$

Steps Using Substitution

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.

$$x+y=360,4\left(x+40\right)+2\left(1.5y-60\right)=1320$$

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

$$x+y=360$$

Subtract y from both sides of the equation.

$$x=-y+360$$

Substitute -y+360 for x in the other equation, 4\left(x+40\right)+2\left(1.5y-60\right)=1320.

$$4\left(-y+360+40\right)+2\left(1.5y-60\right)=1320$$

Add 360 to 40.

$$4\left(-y+400\right)+2\left(1.5y-60\right)=1320$$

Multiply 4 times -y+400.

$$-4y+1600+2\left(1.5y-60\right)=1320$$

Multiply 2 times \frac{3y}{2}-60.

$$-4y+1600+3y-120=1320$$

Add -4y to 3y.

$$-y+1600-120=1320$$

Add 1600 to -120.

$$-y+1480=1320$$

Subtract 1480 from both sides of the equation.

$$-y=-160$$

Divide both sides by -1.

$$y=160$$

Substitute 160 for y in x=-y+360. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=-160+360$$

Add 360 to -160.

$$x=200$$

The system is now solved.

$$x=200,y=160$$

Steps Using Matrices

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

$$x+y=360,4\left(x+40\right)+2\left(1.5y-60\right)=1320$$

Simplify the second equation to put it in standard form.

$$4\left(x+40\right)+2\left(1.5y-60\right)=1320$$

Multiply 4 times x+40.

$$4x+160+2\left(1.5y-60\right)=1320$$

Multiply 2 times 1.5y-60.

$$4x+160+3y-120=1320$$

Add 160 to -120.

$$4x+3y+40=1320$$

Subtract 40 from both sides of the equation.

$$4x+3y=1280$$

Write the equations in matrix form.

$$\left(\begin{matrix}1&1\\4&3\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}360\\1280\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}1&1\\4&3\end{matrix}\right).

$$inverse(\left(\begin{matrix}1&1\\4&3\end{matrix}\right))\left(\begin{matrix}1&1\\4&3\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\4&3\end{matrix}\right))\left(\begin{matrix}360\\1280\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\4&3\end{matrix}\right))\left(\begin{matrix}360\\1280\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\4&3\end{matrix}\right))\left(\begin{matrix}360\\1280\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{3}{3-4}&-\frac{1}{3-4}\\-\frac{4}{3-4}&\frac{1}{3-4}\end{matrix}\right)\left(\begin{matrix}360\\1280\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-3&1\\4&-1\end{matrix}\right)\left(\begin{matrix}360\\1280\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-3\times 360+1280\\4\times 360-1280\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}200\\160\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=200,y=160$$