Solve {c}{19x+19y=40}{3x+8y=15}

$$\left. \begin{array} { c } { 19 x + 19 y = 40 } \\ { 3 x + 8 y = 15 } \end{array} \right.$$

Solve for x, y

$$x=\frac{7}{19}\approx 0.368421053$$

$$y = \frac{33}{19} = 1\frac{14}{19} \approx 1.736842105$$

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.

$$19x+19y=40,3x+8y=15$$

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

$$19x+19y=40$$

Subtract 19y from both sides of the equation.

$$19x=-19y+40$$

Divide both sides by 19.

$$x=\frac{1}{19}\left(-19y+40\right)$$

Multiply \frac{1}{19} times -19y+40.

$$x=-y+\frac{40}{19}$$

Substitute -y+\frac{40}{19} for x in the other equation, 3x+8y=15.

$$3\left(-y+\frac{40}{19}\right)+8y=15$$

Multiply 3 times -y+\frac{40}{19}.

$$-3y+\frac{120}{19}+8y=15$$

Add -3y to 8y.

$$5y+\frac{120}{19}=15$$

Subtract \frac{120}{19} from both sides of the equation.

$$5y=\frac{165}{19}$$

Divide both sides by 5.

$$y=\frac{33}{19}$$

Substitute \frac{33}{19} for y in x=-y+\frac{40}{19}. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=-\frac{33}{19}+\frac{40}{19}$$

Multiply -1 times \frac{33}{19}.

$$x=\frac{-33+40}{19}$$

Add \frac{40}{19} to -\frac{33}{19} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=\frac{7}{19}$$

The system is now solved.

$$x=\frac{7}{19},y=\frac{33}{19}$$

Steps Using Matrices

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

$$19x+19y=40,3x+8y=15$$

Write the equations in matrix form.

$$\left(\begin{matrix}19&19\\3&8\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}40\\15\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}19&19\\3&8\end{matrix}\right).

$$inverse(\left(\begin{matrix}19&19\\3&8\end{matrix}\right))\left(\begin{matrix}19&19\\3&8\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}19&19\\3&8\end{matrix}\right))\left(\begin{matrix}40\\15\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}19&19\\3&8\end{matrix}\right))\left(\begin{matrix}40\\15\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}19&19\\3&8\end{matrix}\right))\left(\begin{matrix}40\\15\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{8}{19\times 8-19\times 3}&-\frac{19}{19\times 8-19\times 3}\\-\frac{3}{19\times 8-19\times 3}&\frac{19}{19\times 8-19\times 3}\end{matrix}\right)\left(\begin{matrix}40\\15\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{8}{95}&-\frac{1}{5}\\-\frac{3}{95}&\frac{1}{5}\end{matrix}\right)\left(\begin{matrix}40\\15\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{8}{95}\times 40-\frac{1}{5}\times 15\\-\frac{3}{95}\times 40+\frac{1}{5}\times 15\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{7}{19}\\\frac{33}{19}\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=\frac{7}{19},y=\frac{33}{19}$$

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.

$$19x+19y=40,3x+8y=15$$

To make 19x and 3x equal, multiply all terms on each side of the first equation by 3 and all terms on each side of the second by 19.

$$3\times 19x+3\times 19y=3\times 40,19\times 3x+19\times 8y=19\times 15$$

Simplify.

$$57x+57y=120,57x+152y=285$$

Subtract 57x+152y=285 from 57x+57y=120 by subtracting like terms on each side of the equal sign.

$$57x-57x+57y-152y=120-285$$

Add 57x to -57x. Terms 57x and -57x cancel out, leaving an equation with only one variable that can be solved.

$$57y-152y=120-285$$

Add 57y to -152y.

$$-95y=120-285$$

Add 120 to -285.

$$-95y=-165$$

Divide both sides by -95.

$$y=\frac{33}{19}$$

Substitute \frac{33}{19} for y in 3x+8y=15. Because the resulting equation contains only one variable, you can solve for x directly.

$$3x+8\times \frac{33}{19}=15$$

Multiply 8 times \frac{33}{19}.

$$3x+\frac{264}{19}=15$$

Subtract \frac{264}{19} from both sides of the equation.

$$3x=\frac{21}{19}$$

Divide both sides by 3.

$$x=\frac{7}{19}$$

The system is now solved.

$$x=\frac{7}{19},y=\frac{33}{19}$$