Solve {l}{2x-5y=100}{4x+y=120}

$$\left. \begin{array} { l } { 2 x - 5 y = 100 } \\ { 4 x + y = 120 } \end{array} \right.$$

Solve for x, y

$$x = \frac{350}{11} = 31\frac{9}{11} \approx 31.818181818$$

$$y = -\frac{80}{11} = -7\frac{3}{11} \approx -7.272727273$$

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.

$$2x-5y=100,4x+y=120$$

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

$$2x-5y=100$$

Add 5y to both sides of the equation.

$$2x=5y+100$$

Divide both sides by 2.

$$x=\frac{1}{2}\left(5y+100\right)$$

Multiply \frac{1}{2} times 100+5y.

$$x=\frac{5}{2}y+50$$

Substitute 50+\frac{5y}{2} for x in the other equation, 4x+y=120.

$$4\left(\frac{5}{2}y+50\right)+y=120$$

Multiply 4 times 50+\frac{5y}{2}.

$$10y+200+y=120$$

Add 10y to y.

$$11y+200=120$$

Subtract 200 from both sides of the equation.

$$11y=-80$$

Divide both sides by 11.

$$y=-\frac{80}{11}$$

Substitute -\frac{80}{11} for y in x=\frac{5}{2}y+50. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=\frac{5}{2}\left(-\frac{80}{11}\right)+50$$

Multiply \frac{5}{2} times -\frac{80}{11} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$x=-\frac{200}{11}+50$$

Add 50 to -\frac{200}{11}.

$$x=\frac{350}{11}$$

The system is now solved.

$$x=\frac{350}{11},y=-\frac{80}{11}$$

Steps Using Matrices

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

$$2x-5y=100,4x+y=120$$

Write the equations in matrix form.

$$\left(\begin{matrix}2&-5\\4&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}100\\120\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}2&-5\\4&1\end{matrix}\right).

$$inverse(\left(\begin{matrix}2&-5\\4&1\end{matrix}\right))\left(\begin{matrix}2&-5\\4&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}2&-5\\4&1\end{matrix}\right))\left(\begin{matrix}100\\120\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}2&-5\\4&1\end{matrix}\right))\left(\begin{matrix}100\\120\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}2&-5\\4&1\end{matrix}\right))\left(\begin{matrix}100\\120\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{2-\left(-5\times 4\right)}&-\frac{-5}{2-\left(-5\times 4\right)}\\-\frac{4}{2-\left(-5\times 4\right)}&\frac{2}{2-\left(-5\times 4\right)}\end{matrix}\right)\left(\begin{matrix}100\\120\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{22}&\frac{5}{22}\\-\frac{2}{11}&\frac{1}{11}\end{matrix}\right)\left(\begin{matrix}100\\120\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{22}\times 100+\frac{5}{22}\times 120\\-\frac{2}{11}\times 100+\frac{1}{11}\times 120\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{350}{11}\\-\frac{80}{11}\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=\frac{350}{11},y=-\frac{80}{11}$$

Steps Using Elimination

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.

$$2x-5y=100,4x+y=120$$

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.

$$4\times 2x+4\left(-5\right)y=4\times 100,2\times 4x+2y=2\times 120$$

Simplify.

$$8x-20y=400,8x+2y=240$$

Subtract 8x+2y=240 from 8x-20y=400 by subtracting like terms on each side of the equal sign.

$$8x-8x-20y-2y=400-240$$

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

$$-20y-2y=400-240$$

Add -20y to -2y.

$$-22y=400-240$$

Add 400 to -240.

$$-22y=160$$

Divide both sides by -22.

$$y=-\frac{80}{11}$$

Substitute -\frac{80}{11} for y in 4x+y=120. Because the resulting equation contains only one variable, you can solve for x directly.

$$4x-\frac{80}{11}=120$$

Add \frac{80}{11} to both sides of the equation.

$$4x=\frac{1400}{11}$$

Divide both sides by 4.

$$x=\frac{350}{11}$$

The system is now solved.

$$x=\frac{350}{11},y=-\frac{80}{11}$$