Solve {l}{340-5x-y=0}{y=15x+200}

$$\left\{ \begin{array} { l } { 340 - 5 x - y = 0 } \\ { y = 15 x + 200 } \end{array} \right.$$

Solve for x, y

$$x=7$$

$$y=305$$

Steps Using Substitution

Consider the first equation. Subtract 340 from both sides. Anything subtracted from zero gives its negation.

$$-5x-y=-340$$

Consider the second equation. Subtract 15x from both sides.

$$y-15x=200$$

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.

$$-5x-y=-340,-15x+y=200$$

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

$$-5x-y=-340$$

Add y to both sides of the equation.

$$-5x=y-340$$

Divide both sides by -5.

$$x=-\frac{1}{5}\left(y-340\right)$$

Multiply -\frac{1}{5} times y-340.

$$x=-\frac{1}{5}y+68$$

Substitute -\frac{y}{5}+68 for x in the other equation, -15x+y=200.

$$-15\left(-\frac{1}{5}y+68\right)+y=200$$

Multiply -15 times -\frac{y}{5}+68.

$$3y-1020+y=200$$

Add 3y to y.

$$4y-1020=200$$

Add 1020 to both sides of the equation.

$$4y=1220$$

Divide both sides by 4.

$$y=305$$

Substitute 305 for y in x=-\frac{1}{5}y+68. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=-\frac{1}{5}\times 305+68$$

Multiply -\frac{1}{5} times 305.

$$x=-61+68$$

Add 68 to -61.

$$x=7$$

The system is now solved.

$$x=7,y=305$$

Steps Using Matrices

Consider the first equation. Subtract 340 from both sides. Anything subtracted from zero gives its negation.

$$-5x-y=-340$$

Consider the second equation. Subtract 15x from both sides.

$$y-15x=200$$

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

$$-5x-y=-340,-15x+y=200$$

Write the equations in matrix form.

$$\left(\begin{matrix}-5&-1\\-15&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-340\\200\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}-5&-1\\-15&1\end{matrix}\right).

$$inverse(\left(\begin{matrix}-5&-1\\-15&1\end{matrix}\right))\left(\begin{matrix}-5&-1\\-15&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}-5&-1\\-15&1\end{matrix}\right))\left(\begin{matrix}-340\\200\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}-5&-1\\-15&1\end{matrix}\right))\left(\begin{matrix}-340\\200\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}-5&-1\\-15&1\end{matrix}\right))\left(\begin{matrix}-340\\200\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{-5-\left(-\left(-15\right)\right)}&-\frac{-1}{-5-\left(-\left(-15\right)\right)}\\-\frac{-15}{-5-\left(-\left(-15\right)\right)}&-\frac{5}{-5-\left(-\left(-15\right)\right)}\end{matrix}\right)\left(\begin{matrix}-340\\200\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{20}&-\frac{1}{20}\\-\frac{3}{4}&\frac{1}{4}\end{matrix}\right)\left(\begin{matrix}-340\\200\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{20}\left(-340\right)-\frac{1}{20}\times 200\\-\frac{3}{4}\left(-340\right)+\frac{1}{4}\times 200\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}7\\305\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=7,y=305$$

Steps Using Elimination

Consider the first equation. Subtract 340 from both sides. Anything subtracted from zero gives its negation.

$$-5x-y=-340$$

Consider the second equation. Subtract 15x from both sides.

$$y-15x=200$$

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.

$$-5x-y=-340,-15x+y=200$$

To make -5x and -15x equal, multiply all terms on each side of the first equation by -15 and all terms on each side of the second by -5.

$$-15\left(-5\right)x-15\left(-1\right)y=-15\left(-340\right),-5\left(-15\right)x-5y=-5\times 200$$

Simplify.

$$75x+15y=5100,75x-5y=-1000$$

Subtract 75x-5y=-1000 from 75x+15y=5100 by subtracting like terms on each side of the equal sign.

$$75x-75x+15y+5y=5100+1000$$

Add 75x to -75x. Terms 75x and -75x cancel out, leaving an equation with only one variable that can be solved.

$$15y+5y=5100+1000$$

Add 15y to 5y.

$$20y=5100+1000$$

Add 5100 to 1000.

$$20y=6100$$

Divide both sides by 20.

$$y=305$$

Substitute 305 for y in -15x+y=200. Because the resulting equation contains only one variable, you can solve for x directly.

$$-15x+305=200$$

Subtract 305 from both sides of the equation.

$$-15x=-105$$

Divide both sides by -15.

$$x=7$$

The system is now solved.

$$x=7,y=305$$