Solve {l}{-10x/3-5y/8=145/12}{-2x/5-frac{3y}{8}=frac{17}{20}}

$$\left. \begin{array} { l } { - \frac { 10 x } { 3 } - \frac { 5 y } { 8 } = \frac { 145 } { 12 } } \\ { - \frac { 2 x } { 5 } - \frac { 3 y } { 8 } = \frac { 17 } { 20 } } \end{array} \right.$$

Solve for x, y

$$x=-4$$

$$y=2$$

Steps Using Substitution

Consider the first equation. Multiply both sides of the equation by 24, the least common multiple of 3,8,12.

$$-8\times 10x-3\times 5y=290$$

Multiply -8 and 10 to get -80.

$$-80x-3\times 5y=290$$

Multiply -3 and 5 to get -15.

$$-80x-15y=290$$

Consider the second equation. Multiply both sides of the equation by 40, the least common multiple of 5,8,20.

$$-8\times 2x-5\times 3y=34$$

Multiply -8 and 2 to get -16.

$$-16x-5\times 3y=34$$

Multiply -5 and 3 to get -15.

$$-16x-15y=34$$

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.

$$-80x-15y=290,-16x-15y=34$$

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

$$-80x-15y=290$$

Add 15y to both sides of the equation.

$$-80x=15y+290$$

Divide both sides by -80.

$$x=-\frac{1}{80}\left(15y+290\right)$$

Multiply -\frac{1}{80} times 15y+290.

$$x=-\frac{3}{16}y-\frac{29}{8}$$

Substitute -\frac{3y}{16}-\frac{29}{8} for x in the other equation, -16x-15y=34.

$$-16\left(-\frac{3}{16}y-\frac{29}{8}\right)-15y=34$$

Multiply -16 times -\frac{3y}{16}-\frac{29}{8}.

$$3y+58-15y=34$$

Add 3y to -15y.

$$-12y+58=34$$

Subtract 58 from both sides of the equation.

$$-12y=-24$$

Divide both sides by -12.

$$y=2$$

Substitute 2 for y in x=-\frac{3}{16}y-\frac{29}{8}. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=-\frac{3}{16}\times 2-\frac{29}{8}$$

Multiply -\frac{3}{16} times 2.

$$x=\frac{-3-29}{8}$$

Add -\frac{29}{8} to -\frac{3}{8} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=-4$$

The system is now solved.

$$x=-4,y=2$$

Steps Using Matrices

Consider the first equation. Multiply both sides of the equation by 24, the least common multiple of 3,8,12.

$$-8\times 10x-3\times 5y=290$$

Multiply -8 and 10 to get -80.

$$-80x-3\times 5y=290$$

Multiply -3 and 5 to get -15.

$$-80x-15y=290$$

Consider the second equation. Multiply both sides of the equation by 40, the least common multiple of 5,8,20.

$$-8\times 2x-5\times 3y=34$$

Multiply -8 and 2 to get -16.

$$-16x-5\times 3y=34$$

Multiply -5 and 3 to get -15.

$$-16x-15y=34$$

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

$$-80x-15y=290,-16x-15y=34$$

Write the equations in matrix form.

$$\left(\begin{matrix}-80&-15\\-16&-15\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}290\\34\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}-80&-15\\-16&-15\end{matrix}\right).

$$inverse(\left(\begin{matrix}-80&-15\\-16&-15\end{matrix}\right))\left(\begin{matrix}-80&-15\\-16&-15\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}-80&-15\\-16&-15\end{matrix}\right))\left(\begin{matrix}290\\34\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}-80&-15\\-16&-15\end{matrix}\right))\left(\begin{matrix}290\\34\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}-80&-15\\-16&-15\end{matrix}\right))\left(\begin{matrix}290\\34\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{15}{-80\left(-15\right)-\left(-15\left(-16\right)\right)}&-\frac{-15}{-80\left(-15\right)-\left(-15\left(-16\right)\right)}\\-\frac{-16}{-80\left(-15\right)-\left(-15\left(-16\right)\right)}&-\frac{80}{-80\left(-15\right)-\left(-15\left(-16\right)\right)}\end{matrix}\right)\left(\begin{matrix}290\\34\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{64}&\frac{1}{64}\\\frac{1}{60}&-\frac{1}{12}\end{matrix}\right)\left(\begin{matrix}290\\34\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{64}\times 290+\frac{1}{64}\times 34\\\frac{1}{60}\times 290-\frac{1}{12}\times 34\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-4\\2\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=-4,y=2$$

Steps Using Elimination

Consider the first equation. Multiply both sides of the equation by 24, the least common multiple of 3,8,12.

$$-8\times 10x-3\times 5y=290$$

Multiply -8 and 10 to get -80.

$$-80x-3\times 5y=290$$

Multiply -3 and 5 to get -15.

$$-80x-15y=290$$

Consider the second equation. Multiply both sides of the equation by 40, the least common multiple of 5,8,20.

$$-8\times 2x-5\times 3y=34$$

Multiply -8 and 2 to get -16.

$$-16x-5\times 3y=34$$

Multiply -5 and 3 to get -15.

$$-16x-15y=34$$

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.

$$-80x-15y=290,-16x-15y=34$$

Subtract -16x-15y=34 from -80x-15y=290 by subtracting like terms on each side of the equal sign.

$$-80x+16x-15y+15y=290-34$$

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

$$-80x+16x=290-34$$

Add -80x to 16x.

$$-64x=290-34$$

Add 290 to -34.

$$-64x=256$$

Divide both sides by -64.

$$x=-4$$

Substitute -4 for x in -16x-15y=34. Because the resulting equation contains only one variable, you can solve for y directly.

$$-16\left(-4\right)-15y=34$$

Multiply -16 times -4.

$$64-15y=34$$

Subtract 64 from both sides of the equation.

$$-15y=-30$$

Divide both sides by -15.

$$y=2$$

The system is now solved.

$$x=-4,y=2$$