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

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

Solve for x, y

$$x=4$$

$$y=0$$

Steps Using Substitution

Consider the first equation. Multiply both sides of the equation by 20, the least common multiple of 2,5,4.

$$10\left(x+2\right)+4\left(y-5\right)=5x+20$$

Use the distributive property to multiply 10 by x+2.

$$10x+20+4\left(y-5\right)=5x+20$$

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

$$10x+20+4y-20=5x+20$$

Subtract 20 from 20 to get 0.

$$10x+4y=5x+20$$

Subtract 5x from both sides.

$$10x+4y-5x=20$$

Combine 10x and -5x to get 5x.

$$5x+4y=20$$

Consider the second equation. Multiply both sides of the equation by 3.

$$3x+3y=x-1+9$$

Add -1 and 9 to get 8.

$$3x+3y=x+8$$

Subtract x from both sides.

$$3x+3y-x=8$$

Combine 3x and -x to get 2x.

$$2x+3y=8$$

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.

$$5x+4y=20,2x+3y=8$$

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

$$5x+4y=20$$

Subtract 4y from both sides of the equation.

$$5x=-4y+20$$

Divide both sides by 5.

$$x=\frac{1}{5}\left(-4y+20\right)$$

Multiply \frac{1}{5} times -4y+20.

$$x=-\frac{4}{5}y+4$$

Substitute -\frac{4y}{5}+4 for x in the other equation, 2x+3y=8.

$$2\left(-\frac{4}{5}y+4\right)+3y=8$$

Multiply 2 times -\frac{4y}{5}+4.

$$-\frac{8}{5}y+8+3y=8$$

Add -\frac{8y}{5} to 3y.

$$\frac{7}{5}y+8=8$$

Subtract 8 from both sides of the equation.

$$\frac{7}{5}y=0$$

Divide both sides of the equation by \frac{7}{5}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$y=0$$

Substitute 0 for y in x=-\frac{4}{5}y+4. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=4$$

The system is now solved.

$$x=4,y=0$$

Steps Using Matrices

Consider the first equation. Multiply both sides of the equation by 20, the least common multiple of 2,5,4.

$$10\left(x+2\right)+4\left(y-5\right)=5x+20$$

Use the distributive property to multiply 10 by x+2.

$$10x+20+4\left(y-5\right)=5x+20$$

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

$$10x+20+4y-20=5x+20$$

Subtract 20 from 20 to get 0.

$$10x+4y=5x+20$$

Subtract 5x from both sides.

$$10x+4y-5x=20$$

Combine 10x and -5x to get 5x.

$$5x+4y=20$$

Consider the second equation. Multiply both sides of the equation by 3.

$$3x+3y=x-1+9$$

Add -1 and 9 to get 8.

$$3x+3y=x+8$$

Subtract x from both sides.

$$3x+3y-x=8$$

Combine 3x and -x to get 2x.

$$2x+3y=8$$

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

$$5x+4y=20,2x+3y=8$$

Write the equations in matrix form.

$$\left(\begin{matrix}5&4\\2&3\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}20\\8\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}5&4\\2&3\end{matrix}\right).

$$inverse(\left(\begin{matrix}5&4\\2&3\end{matrix}\right))\left(\begin{matrix}5&4\\2&3\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}5&4\\2&3\end{matrix}\right))\left(\begin{matrix}20\\8\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}5&4\\2&3\end{matrix}\right))\left(\begin{matrix}20\\8\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}5&4\\2&3\end{matrix}\right))\left(\begin{matrix}20\\8\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{3}{5\times 3-4\times 2}&-\frac{4}{5\times 3-4\times 2}\\-\frac{2}{5\times 3-4\times 2}&\frac{5}{5\times 3-4\times 2}\end{matrix}\right)\left(\begin{matrix}20\\8\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{3}{7}&-\frac{4}{7}\\-\frac{2}{7}&\frac{5}{7}\end{matrix}\right)\left(\begin{matrix}20\\8\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{3}{7}\times 20-\frac{4}{7}\times 8\\-\frac{2}{7}\times 20+\frac{5}{7}\times 8\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}4\\0\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=4,y=0$$

Steps Using Elimination

Consider the first equation. Multiply both sides of the equation by 20, the least common multiple of 2,5,4.

$$10\left(x+2\right)+4\left(y-5\right)=5x+20$$

Use the distributive property to multiply 10 by x+2.

$$10x+20+4\left(y-5\right)=5x+20$$

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

$$10x+20+4y-20=5x+20$$

Subtract 20 from 20 to get 0.

$$10x+4y=5x+20$$

Subtract 5x from both sides.

$$10x+4y-5x=20$$

Combine 10x and -5x to get 5x.

$$5x+4y=20$$

Consider the second equation. Multiply both sides of the equation by 3.

$$3x+3y=x-1+9$$

Add -1 and 9 to get 8.

$$3x+3y=x+8$$

Subtract x from both sides.

$$3x+3y-x=8$$

Combine 3x and -x to get 2x.

$$2x+3y=8$$

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.

$$5x+4y=20,2x+3y=8$$

To make 5x and 2x equal, multiply all terms on each side of the first equation by 2 and all terms on each side of the second by 5.

$$2\times 5x+2\times 4y=2\times 20,5\times 2x+5\times 3y=5\times 8$$

Simplify.

$$10x+8y=40,10x+15y=40$$

Subtract 10x+15y=40 from 10x+8y=40 by subtracting like terms on each side of the equal sign.

$$10x-10x+8y-15y=40-40$$

Add 10x to -10x. Terms 10x and -10x cancel out, leaving an equation with only one variable that can be solved.

$$8y-15y=40-40$$

Add 8y to -15y.

$$-7y=40-40$$

Add 40 to -40.

$$-7y=0$$

Divide both sides by -7.

$$y=0$$

Substitute 0 for y in 2x+3y=8. Because the resulting equation contains only one variable, you can solve for x directly.

$$2x=8$$

Divide both sides by 2.

$$x=4$$

The system is now solved.

$$x=4,y=0$$