Solve {l}{x+7/5-2x-y/4=2-5y}{5y-4/2+4x-7/3=3x-1}

$$\left. \begin{array} { l } { \frac { x + 7 } { 5 } - \frac { 2 x - y } { 4 } = 2 - 5 y } \\ { \frac { 5 y - 4 } { 2 } + \frac { 4 x - 7 } { 3 } = 3 x - 1 } \end{array} \right.$$

Solve for x, y

$$x=-2$$

$$y=0$$

Steps Using Substitution

Consider the first equation. Multiply both sides of the equation by 20, the least common multiple of 5,4.

$$4\left(x+7\right)-5\left(2x-y\right)=40-100y$$

Use the distributive property to multiply 4 by x+7.

$$4x+28-5\left(2x-y\right)=40-100y$$

Use the distributive property to multiply -5 by 2x-y.

$$4x+28-10x+5y=40-100y$$

Combine 4x and -10x to get -6x.

$$-6x+28+5y=40-100y$$

Add 100y to both sides.

$$-6x+28+5y+100y=40$$

Combine 5y and 100y to get 105y.

$$-6x+28+105y=40$$

Subtract 28 from both sides.

$$-6x+105y=40-28$$

Subtract 28 from 40 to get 12.

$$-6x+105y=12$$

Consider the second equation. Multiply both sides of the equation by 6, the least common multiple of 2,3.

$$3\left(5y-4\right)+2\left(4x-7\right)=18x-6$$

Use the distributive property to multiply 3 by 5y-4.

$$15y-12+2\left(4x-7\right)=18x-6$$

Use the distributive property to multiply 2 by 4x-7.

$$15y-12+8x-14=18x-6$$

Subtract 14 from -12 to get -26.

$$15y-26+8x=18x-6$$

Subtract 18x from both sides.

$$15y-26+8x-18x=-6$$

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

$$15y-26-10x=-6$$

Add 26 to both sides.

$$15y-10x=-6+26$$

Add -6 and 26 to get 20.

$$15y-10x=20$$

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.

$$-6x+105y=12,-10x+15y=20$$

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

$$-6x+105y=12$$

Subtract 105y from both sides of the equation.

$$-6x=-105y+12$$

Divide both sides by -6.

$$x=-\frac{1}{6}\left(-105y+12\right)$$

Multiply -\frac{1}{6} times -105y+12.

$$x=\frac{35}{2}y-2$$

Substitute \frac{35y}{2}-2 for x in the other equation, -10x+15y=20.

$$-10\left(\frac{35}{2}y-2\right)+15y=20$$

Multiply -10 times \frac{35y}{2}-2.

$$-175y+20+15y=20$$

Add -175y to 15y.

$$-160y+20=20$$

Subtract 20 from both sides of the equation.

$$-160y=0$$

Divide both sides by -160.

$$y=0$$

Substitute 0 for y in x=\frac{35}{2}y-2. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=-2$$

The system is now solved.

$$x=-2,y=0$$

Steps Using Matrices

Consider the first equation. Multiply both sides of the equation by 20, the least common multiple of 5,4.

$$4\left(x+7\right)-5\left(2x-y\right)=40-100y$$

Use the distributive property to multiply 4 by x+7.

$$4x+28-5\left(2x-y\right)=40-100y$$

Use the distributive property to multiply -5 by 2x-y.

$$4x+28-10x+5y=40-100y$$

Combine 4x and -10x to get -6x.

$$-6x+28+5y=40-100y$$

Add 100y to both sides.

$$-6x+28+5y+100y=40$$

Combine 5y and 100y to get 105y.

$$-6x+28+105y=40$$

Subtract 28 from both sides.

$$-6x+105y=40-28$$

Subtract 28 from 40 to get 12.

$$-6x+105y=12$$

Consider the second equation. Multiply both sides of the equation by 6, the least common multiple of 2,3.

$$3\left(5y-4\right)+2\left(4x-7\right)=18x-6$$

Use the distributive property to multiply 3 by 5y-4.

$$15y-12+2\left(4x-7\right)=18x-6$$

Use the distributive property to multiply 2 by 4x-7.

$$15y-12+8x-14=18x-6$$

Subtract 14 from -12 to get -26.

$$15y-26+8x=18x-6$$

Subtract 18x from both sides.

$$15y-26+8x-18x=-6$$

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

$$15y-26-10x=-6$$

Add 26 to both sides.

$$15y-10x=-6+26$$

Add -6 and 26 to get 20.

$$15y-10x=20$$

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

$$-6x+105y=12,-10x+15y=20$$

Write the equations in matrix form.

$$\left(\begin{matrix}-6&105\\-10&15\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}12\\20\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}-6&105\\-10&15\end{matrix}\right).

$$inverse(\left(\begin{matrix}-6&105\\-10&15\end{matrix}\right))\left(\begin{matrix}-6&105\\-10&15\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}-6&105\\-10&15\end{matrix}\right))\left(\begin{matrix}12\\20\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}-6&105\\-10&15\end{matrix}\right))\left(\begin{matrix}12\\20\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}-6&105\\-10&15\end{matrix}\right))\left(\begin{matrix}12\\20\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{15}{-6\times 15-105\left(-10\right)}&-\frac{105}{-6\times 15-105\left(-10\right)}\\-\frac{-10}{-6\times 15-105\left(-10\right)}&-\frac{6}{-6\times 15-105\left(-10\right)}\end{matrix}\right)\left(\begin{matrix}12\\20\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{64}&-\frac{7}{64}\\\frac{1}{96}&-\frac{1}{160}\end{matrix}\right)\left(\begin{matrix}12\\20\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{64}\times 12-\frac{7}{64}\times 20\\\frac{1}{96}\times 12-\frac{1}{160}\times 20\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-2\\0\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=-2,y=0$$

Steps Using Elimination

Consider the first equation. Multiply both sides of the equation by 20, the least common multiple of 5,4.

$$4\left(x+7\right)-5\left(2x-y\right)=40-100y$$

Use the distributive property to multiply 4 by x+7.

$$4x+28-5\left(2x-y\right)=40-100y$$

Use the distributive property to multiply -5 by 2x-y.

$$4x+28-10x+5y=40-100y$$

Combine 4x and -10x to get -6x.

$$-6x+28+5y=40-100y$$

Add 100y to both sides.

$$-6x+28+5y+100y=40$$

Combine 5y and 100y to get 105y.

$$-6x+28+105y=40$$

Subtract 28 from both sides.

$$-6x+105y=40-28$$

Subtract 28 from 40 to get 12.

$$-6x+105y=12$$

Consider the second equation. Multiply both sides of the equation by 6, the least common multiple of 2,3.

$$3\left(5y-4\right)+2\left(4x-7\right)=18x-6$$

Use the distributive property to multiply 3 by 5y-4.

$$15y-12+2\left(4x-7\right)=18x-6$$

Use the distributive property to multiply 2 by 4x-7.

$$15y-12+8x-14=18x-6$$

Subtract 14 from -12 to get -26.

$$15y-26+8x=18x-6$$

Subtract 18x from both sides.

$$15y-26+8x-18x=-6$$

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

$$15y-26-10x=-6$$

Add 26 to both sides.

$$15y-10x=-6+26$$

Add -6 and 26 to get 20.

$$15y-10x=20$$

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.

$$-6x+105y=12,-10x+15y=20$$

To make -6x and -10x equal, multiply all terms on each side of the first equation by -10 and all terms on each side of the second by -6.

$$-10\left(-6\right)x-10\times 105y=-10\times 12,-6\left(-10\right)x-6\times 15y=-6\times 20$$

Simplify.

$$60x-1050y=-120,60x-90y=-120$$

Subtract 60x-90y=-120 from 60x-1050y=-120 by subtracting like terms on each side of the equal sign.

$$60x-60x-1050y+90y=-120+120$$

Add 60x to -60x. Terms 60x and -60x cancel out, leaving an equation with only one variable that can be solved.

$$-1050y+90y=-120+120$$

Add -1050y to 90y.

$$-960y=-120+120$$

Add -120 to 120.

$$-960y=0$$

Divide both sides by -960.

$$y=0$$

Substitute 0 for y in -10x+15y=20. Because the resulting equation contains only one variable, you can solve for x directly.

$$-10x=20$$

Divide both sides by -10.

$$x=-2$$

The system is now solved.

$$x=-2,y=0$$