Solve {l}{x+y+3x+6/2=72}{x+y/6=10}

$$\left\{ \begin{array} { l } { \frac { x + y + 3 x + 6 } { 2 } = 72 } \\ { \frac { x + y } { 6 } = 10 } \end{array} \right.$$

Solve for x, y

$$x=26$$

$$y=34$$

Steps Using Substitution

Consider the first equation. Multiply both sides by 2.

$$x+y+3x+6=72\times 2$$

Combine x and 3x to get 4x.

$$4x+y+6=72\times 2$$

Multiply 72 and 2 to get 144.

$$4x+y+6=144$$

Subtract 6 from both sides.

$$4x+y=144-6$$

Subtract 6 from 144 to get 138.

$$4x+y=138$$

Consider the second equation. Multiply both sides by 6.

$$x+y=10\times 6$$

Multiply 10 and 6 to get 60.

$$x+y=60$$

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.

$$4x+y=138,x+y=60$$

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

$$4x+y=138$$

Subtract y from both sides of the equation.

$$4x=-y+138$$

Divide both sides by 4.

$$x=\frac{1}{4}\left(-y+138\right)$$

Multiply \frac{1}{4} times -y+138.

$$x=-\frac{1}{4}y+\frac{69}{2}$$

Substitute -\frac{y}{4}+\frac{69}{2} for x in the other equation, x+y=60.

$$-\frac{1}{4}y+\frac{69}{2}+y=60$$

Add -\frac{y}{4} to y.

$$\frac{3}{4}y+\frac{69}{2}=60$$

Subtract \frac{69}{2} from both sides of the equation.

$$\frac{3}{4}y=\frac{51}{2}$$

Divide both sides of the equation by \frac{3}{4}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$y=34$$

Substitute 34 for y in x=-\frac{1}{4}y+\frac{69}{2}. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=-\frac{1}{4}\times 34+\frac{69}{2}$$

Multiply -\frac{1}{4} times 34.

$$x=\frac{-17+69}{2}$$

Add \frac{69}{2} to -\frac{17}{2} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=26$$

The system is now solved.

$$x=26,y=34$$

Steps Using Matrices

Consider the first equation. Multiply both sides by 2.

$$x+y+3x+6=72\times 2$$

Combine x and 3x to get 4x.

$$4x+y+6=72\times 2$$

Multiply 72 and 2 to get 144.

$$4x+y+6=144$$

Subtract 6 from both sides.

$$4x+y=144-6$$

Subtract 6 from 144 to get 138.

$$4x+y=138$$

Consider the second equation. Multiply both sides by 6.

$$x+y=10\times 6$$

Multiply 10 and 6 to get 60.

$$x+y=60$$

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

$$4x+y=138,x+y=60$$

Write the equations in matrix form.

$$\left(\begin{matrix}4&1\\1&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}138\\60\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}4&1\\1&1\end{matrix}\right).

$$inverse(\left(\begin{matrix}4&1\\1&1\end{matrix}\right))\left(\begin{matrix}4&1\\1&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}4&1\\1&1\end{matrix}\right))\left(\begin{matrix}138\\60\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}4&1\\1&1\end{matrix}\right))\left(\begin{matrix}138\\60\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}4&1\\1&1\end{matrix}\right))\left(\begin{matrix}138\\60\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{4-1}&-\frac{1}{4-1}\\-\frac{1}{4-1}&\frac{4}{4-1}\end{matrix}\right)\left(\begin{matrix}138\\60\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{3}&-\frac{1}{3}\\-\frac{1}{3}&\frac{4}{3}\end{matrix}\right)\left(\begin{matrix}138\\60\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{3}\times 138-\frac{1}{3}\times 60\\-\frac{1}{3}\times 138+\frac{4}{3}\times 60\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}26\\34\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=26,y=34$$

Steps Using Elimination

Consider the first equation. Multiply both sides by 2.

$$x+y+3x+6=72\times 2$$

Combine x and 3x to get 4x.

$$4x+y+6=72\times 2$$

Multiply 72 and 2 to get 144.

$$4x+y+6=144$$

Subtract 6 from both sides.

$$4x+y=144-6$$

Subtract 6 from 144 to get 138.

$$4x+y=138$$

Consider the second equation. Multiply both sides by 6.

$$x+y=10\times 6$$

Multiply 10 and 6 to get 60.

$$x+y=60$$

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.

$$4x+y=138,x+y=60$$

Subtract x+y=60 from 4x+y=138 by subtracting like terms on each side of the equal sign.

$$4x-x+y-y=138-60$$

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

$$4x-x=138-60$$

Add 4x to -x.

$$3x=138-60$$

Add 138 to -60.

$$3x=78$$

Divide both sides by 3.

$$x=26$$

Substitute 26 for x in x+y=60. Because the resulting equation contains only one variable, you can solve for y directly.

$$26+y=60$$

Subtract 26 from both sides of the equation.

$$y=34$$

The system is now solved.

$$x=26,y=34$$