Solve {r}{x/2-y/4=2}{x+y/5-2x-y/4=1-frac{x}{6}}

$$\left. \begin{array} { r } { \frac { x } { 2 } - \frac { y } { 4 } = 2 } \\ { \frac { x + y } { 5 } - \frac { 2 x - y } { 4 } = 1 - \frac { x } { 6 } } \end{array} \right.$$

Solve for x, y

$$x=6$$

$$y=4$$

Steps Using Substitution

Consider the first equation. Multiply both sides of the equation by 4, the least common multiple of 2,4.

$$2x-y=8$$

Consider the second equation. Multiply both sides of the equation by 60, the least common multiple of 5,4,6.

$$12\left(x+y\right)-15\left(2x-y\right)=60-10x$$

Use the distributive property to multiply 12 by x+y.

$$12x+12y-15\left(2x-y\right)=60-10x$$

Use the distributive property to multiply -15 by 2x-y.

$$12x+12y-30x+15y=60-10x$$

Combine 12x and -30x to get -18x.

$$-18x+12y+15y=60-10x$$

Combine 12y and 15y to get 27y.

$$-18x+27y=60-10x$$

Add 10x to both sides.

$$-18x+27y+10x=60$$

Combine -18x and 10x to get -8x.

$$-8x+27y=60$$

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.

$$2x-y=8,-8x+27y=60$$

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

$$2x-y=8$$

Add y to both sides of the equation.

$$2x=y+8$$

Divide both sides by 2.

$$x=\frac{1}{2}\left(y+8\right)$$

Multiply \frac{1}{2} times y+8.

$$x=\frac{1}{2}y+4$$

Substitute \frac{y}{2}+4 for x in the other equation, -8x+27y=60.

$$-8\left(\frac{1}{2}y+4\right)+27y=60$$

Multiply -8 times \frac{y}{2}+4.

$$-4y-32+27y=60$$

Add -4y to 27y.

$$23y-32=60$$

Add 32 to both sides of the equation.

$$23y=92$$

Divide both sides by 23.

$$y=4$$

Substitute 4 for y in x=\frac{1}{2}y+4. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=\frac{1}{2}\times 4+4$$

Multiply \frac{1}{2} times 4.

$$x=2+4$$

Add 4 to 2.

$$x=6$$

The system is now solved.

$$x=6,y=4$$

Steps Using Matrices

Consider the first equation. Multiply both sides of the equation by 4, the least common multiple of 2,4.

$$2x-y=8$$

Consider the second equation. Multiply both sides of the equation by 60, the least common multiple of 5,4,6.

$$12\left(x+y\right)-15\left(2x-y\right)=60-10x$$

Use the distributive property to multiply 12 by x+y.

$$12x+12y-15\left(2x-y\right)=60-10x$$

Use the distributive property to multiply -15 by 2x-y.

$$12x+12y-30x+15y=60-10x$$

Combine 12x and -30x to get -18x.

$$-18x+12y+15y=60-10x$$

Combine 12y and 15y to get 27y.

$$-18x+27y=60-10x$$

Add 10x to both sides.

$$-18x+27y+10x=60$$

Combine -18x and 10x to get -8x.

$$-8x+27y=60$$

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

$$2x-y=8,-8x+27y=60$$

Write the equations in matrix form.

$$\left(\begin{matrix}2&-1\\-8&27\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}8\\60\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}2&-1\\-8&27\end{matrix}\right).

$$inverse(\left(\begin{matrix}2&-1\\-8&27\end{matrix}\right))\left(\begin{matrix}2&-1\\-8&27\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}2&-1\\-8&27\end{matrix}\right))\left(\begin{matrix}8\\60\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}2&-1\\-8&27\end{matrix}\right))\left(\begin{matrix}8\\60\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}2&-1\\-8&27\end{matrix}\right))\left(\begin{matrix}8\\60\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{27}{2\times 27-\left(-\left(-8\right)\right)}&-\frac{-1}{2\times 27-\left(-\left(-8\right)\right)}\\-\frac{-8}{2\times 27-\left(-\left(-8\right)\right)}&\frac{2}{2\times 27-\left(-\left(-8\right)\right)}\end{matrix}\right)\left(\begin{matrix}8\\60\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{27}{46}&\frac{1}{46}\\\frac{4}{23}&\frac{1}{23}\end{matrix}\right)\left(\begin{matrix}8\\60\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{27}{46}\times 8+\frac{1}{46}\times 60\\\frac{4}{23}\times 8+\frac{1}{23}\times 60\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}6\\4\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=6,y=4$$

Steps Using Elimination

Consider the first equation. Multiply both sides of the equation by 4, the least common multiple of 2,4.

$$2x-y=8$$

Consider the second equation. Multiply both sides of the equation by 60, the least common multiple of 5,4,6.

$$12\left(x+y\right)-15\left(2x-y\right)=60-10x$$

Use the distributive property to multiply 12 by x+y.

$$12x+12y-15\left(2x-y\right)=60-10x$$

Use the distributive property to multiply -15 by 2x-y.

$$12x+12y-30x+15y=60-10x$$

Combine 12x and -30x to get -18x.

$$-18x+12y+15y=60-10x$$

Combine 12y and 15y to get 27y.

$$-18x+27y=60-10x$$

Add 10x to both sides.

$$-18x+27y+10x=60$$

Combine -18x and 10x to get -8x.

$$-8x+27y=60$$

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.

$$2x-y=8,-8x+27y=60$$

To make 2x and -8x equal, multiply all terms on each side of the first equation by -8 and all terms on each side of the second by 2.

$$-8\times 2x-8\left(-1\right)y=-8\times 8,2\left(-8\right)x+2\times 27y=2\times 60$$

Simplify.

$$-16x+8y=-64,-16x+54y=120$$

Subtract -16x+54y=120 from -16x+8y=-64 by subtracting like terms on each side of the equal sign.

$$-16x+16x+8y-54y=-64-120$$

Add -16x to 16x. Terms -16x and 16x cancel out, leaving an equation with only one variable that can be solved.

$$8y-54y=-64-120$$

Add 8y to -54y.

$$-46y=-64-120$$

Add -64 to -120.

$$-46y=-184$$

Divide both sides by -46.

$$y=4$$

Substitute 4 for y in -8x+27y=60. Because the resulting equation contains only one variable, you can solve for x directly.

$$-8x+27\times 4=60$$

Multiply 27 times 4.

$$-8x+108=60$$

Subtract 108 from both sides of the equation.

$$-8x=-48$$

Divide both sides by -8.

$$x=6$$

The system is now solved.

$$x=6,y=4$$