Solve {l}{1.8+220x+60y=316}{108+220x+100y=352}

$$\left. \begin{array} { l } { 1.8 + 220 x + 60 y = 316 } \\ { 108 + 220 x + 100 y = 352 } \end{array} \right.$$

Solve for x, y

$$x = \frac{839}{440} = 1\frac{399}{440} \approx 1.906818182$$

$$y=-1.755$$

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.

$$220x+60y+1.8=316,220x+100y+108=352$$

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

$$220x+60y+1.8=316$$

Subtract 1.8 from both sides of the equation.

$$220x+60y=314.2$$

Subtract 60y from both sides of the equation.

$$220x=-60y+314.2$$

Divide both sides by 220.

$$x=\frac{1}{220}\left(-60y+314.2\right)$$

Multiply \frac{1}{220} times -60y+314.2.

$$x=-\frac{3}{11}y+\frac{1571}{1100}$$

Substitute -\frac{3y}{11}+\frac{1571}{1100} for x in the other equation, 220x+100y+108=352.

$$220\left(-\frac{3}{11}y+\frac{1571}{1100}\right)+100y+108=352$$

Multiply 220 times -\frac{3y}{11}+\frac{1571}{1100}.

$$-60y+\frac{1571}{5}+100y+108=352$$

Add -60y to 100y.

$$40y+\frac{1571}{5}+108=352$$

Add \frac{1571}{5} to 108.

$$40y+\frac{2111}{5}=352$$

Subtract \frac{2111}{5} from both sides of the equation.

$$40y=-\frac{351}{5}$$

Divide both sides by 40.

$$y=-\frac{351}{200}$$

Substitute -\frac{351}{200} for y in x=-\frac{3}{11}y+\frac{1571}{1100}. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=-\frac{3}{11}\left(-\frac{351}{200}\right)+\frac{1571}{1100}$$

Multiply -\frac{3}{11} times -\frac{351}{200} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$x=\frac{1053}{2200}+\frac{1571}{1100}$$

Add \frac{1571}{1100} to \frac{1053}{2200} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=\frac{839}{440}$$

The system is now solved.

$$x=\frac{839}{440},y=-\frac{351}{200}$$

Steps Using Matrices

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

$$220x+60y+1.8=316,220x+100y+108=352$$

Write the equations in matrix form.

$$\left(\begin{matrix}220&60\\220&100\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}314.2\\244\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}220&60\\220&100\end{matrix}\right).

$$inverse(\left(\begin{matrix}220&60\\220&100\end{matrix}\right))\left(\begin{matrix}220&60\\220&100\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}220&60\\220&100\end{matrix}\right))\left(\begin{matrix}314.2\\244\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}220&60\\220&100\end{matrix}\right))\left(\begin{matrix}314.2\\244\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}220&60\\220&100\end{matrix}\right))\left(\begin{matrix}314.2\\244\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{100}{220\times 100-60\times 220}&-\frac{60}{220\times 100-60\times 220}\\-\frac{220}{220\times 100-60\times 220}&\frac{220}{220\times 100-60\times 220}\end{matrix}\right)\left(\begin{matrix}314.2\\244\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{88}&-\frac{3}{440}\\-\frac{1}{40}&\frac{1}{40}\end{matrix}\right)\left(\begin{matrix}314.2\\244\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{88}\times 314.2-\frac{3}{440}\times 244\\-\frac{1}{40}\times 314.2+\frac{1}{40}\times 244\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{839}{440}\\-\frac{351}{200}\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=\frac{839}{440},y=-\frac{351}{200}$$

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.

$$220x+60y+1.8=316,220x+100y+108=352$$

Subtract 220x+100y+108=352 from 220x+60y+1.8=316 by subtracting like terms on each side of the equal sign.

$$220x-220x+60y-100y+1.8-108=316-352$$

Add 220x to -220x. Terms 220x and -220x cancel out, leaving an equation with only one variable that can be solved.

$$60y-100y+1.8-108=316-352$$

Add 60y to -100y.

$$-40y+1.8-108=316-352$$

Add 1.8 to -108.

$$-40y-106.2=316-352$$

Add 316 to -352.

$$-40y-106.2=-36$$

Add 106.2 to both sides of the equation.

$$-40y=70.2$$

Divide both sides by -40.

$$y=-\frac{351}{200}$$

Substitute -\frac{351}{200} for y in 220x+100y+108=352. Because the resulting equation contains only one variable, you can solve for x directly.

$$220x+100\left(-\frac{351}{200}\right)+108=352$$

Multiply 100 times -\frac{351}{200}.

$$220x-\frac{351}{2}+108=352$$

Add -\frac{351}{2} to 108.

$$220x-\frac{135}{2}=352$$

Add \frac{135}{2} to both sides of the equation.

$$220x=\frac{839}{2}$$

Divide both sides by 220.

$$x=\frac{839}{440}$$

The system is now solved.

$$x=\frac{839}{440},y=-\frac{351}{200}$$