Solve {l}{3x+9y=34}{5x+35y=99}

$$\left. \begin{array} { l } { 3 x + 9 y = 34 } \\ { 5 x + 35 y = 99 } \end{array} \right.$$

Solve for x, y

$$x = \frac{299}{60} = 4\frac{59}{60} \approx 4.983333333$$

$$y = \frac{127}{60} = 2\frac{7}{60} \approx 2.116666667$$

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.

$$3x+9y=34,5x+35y=99$$

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

$$3x+9y=34$$

Subtract 9y from both sides of the equation.

$$3x=-9y+34$$

Divide both sides by 3.

$$x=\frac{1}{3}\left(-9y+34\right)$$

Multiply \frac{1}{3} times -9y+34.

$$x=-3y+\frac{34}{3}$$

Substitute -3y+\frac{34}{3} for x in the other equation, 5x+35y=99.

$$5\left(-3y+\frac{34}{3}\right)+35y=99$$

Multiply 5 times -3y+\frac{34}{3}.

$$-15y+\frac{170}{3}+35y=99$$

Add -15y to 35y.

$$20y+\frac{170}{3}=99$$

Subtract \frac{170}{3} from both sides of the equation.

$$20y=\frac{127}{3}$$

Divide both sides by 20.

$$y=\frac{127}{60}$$

Substitute \frac{127}{60} for y in x=-3y+\frac{34}{3}. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=-3\times \frac{127}{60}+\frac{34}{3}$$

Multiply -3 times \frac{127}{60}.

$$x=-\frac{127}{20}+\frac{34}{3}$$

Add \frac{34}{3} to -\frac{127}{20} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=\frac{299}{60}$$

The system is now solved.

$$x=\frac{299}{60},y=\frac{127}{60}$$

Steps Using Matrices

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

$$3x+9y=34,5x+35y=99$$

Write the equations in matrix form.

$$\left(\begin{matrix}3&9\\5&35\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}34\\99\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}3&9\\5&35\end{matrix}\right).

$$inverse(\left(\begin{matrix}3&9\\5&35\end{matrix}\right))\left(\begin{matrix}3&9\\5&35\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}3&9\\5&35\end{matrix}\right))\left(\begin{matrix}34\\99\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}3&9\\5&35\end{matrix}\right))\left(\begin{matrix}34\\99\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}3&9\\5&35\end{matrix}\right))\left(\begin{matrix}34\\99\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{35}{3\times 35-9\times 5}&-\frac{9}{3\times 35-9\times 5}\\-\frac{5}{3\times 35-9\times 5}&\frac{3}{3\times 35-9\times 5}\end{matrix}\right)\left(\begin{matrix}34\\99\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{7}{12}&-\frac{3}{20}\\-\frac{1}{12}&\frac{1}{20}\end{matrix}\right)\left(\begin{matrix}34\\99\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{7}{12}\times 34-\frac{3}{20}\times 99\\-\frac{1}{12}\times 34+\frac{1}{20}\times 99\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{299}{60}\\\frac{127}{60}\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=\frac{299}{60},y=\frac{127}{60}$$

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.

$$3x+9y=34,5x+35y=99$$

To make 3x and 5x equal, multiply all terms on each side of the first equation by 5 and all terms on each side of the second by 3.

$$5\times 3x+5\times 9y=5\times 34,3\times 5x+3\times 35y=3\times 99$$

Simplify.

$$15x+45y=170,15x+105y=297$$

Subtract 15x+105y=297 from 15x+45y=170 by subtracting like terms on each side of the equal sign.

$$15x-15x+45y-105y=170-297$$

Add 15x to -15x. Terms 15x and -15x cancel out, leaving an equation with only one variable that can be solved.

$$45y-105y=170-297$$

Add 45y to -105y.

$$-60y=170-297$$

Add 170 to -297.

$$-60y=-127$$

Divide both sides by -60.

$$y=\frac{127}{60}$$

Substitute \frac{127}{60} for y in 5x+35y=99. Because the resulting equation contains only one variable, you can solve for x directly.

$$5x+35\times \frac{127}{60}=99$$

Multiply 35 times \frac{127}{60}.

$$5x+\frac{889}{12}=99$$

Subtract \frac{889}{12} from both sides of the equation.

$$5x=\frac{299}{12}$$

Divide both sides by 5.

$$x=\frac{299}{60}$$

The system is now solved.

$$x=\frac{299}{60},y=\frac{127}{60}$$