Solve {l}{x/6+y/5=5}{x/4-y/3=-2}

$$\left. \begin{array} { l } { \frac { x } { 6 } + \frac { y } { 5 } = 5 } \\ { \frac { x } { 4 } - \frac { y } { 3 } = - 2 } \end{array} \right.$$

Solve for x, y

$$x=12$$

$$y=15$$

Steps Using Substitution

Consider the first equation. Multiply both sides of the equation by 30, the least common multiple of 6,5.

$$5x+6y=150$$

Consider the second equation. Multiply both sides of the equation by 12, the least common multiple of 4,3.

$$3x-4y=-24$$

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.

$$5x+6y=150,3x-4y=-24$$

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

$$5x+6y=150$$

Subtract 6y from both sides of the equation.

$$5x=-6y+150$$

Divide both sides by 5.

$$x=\frac{1}{5}\left(-6y+150\right)$$

Multiply \frac{1}{5} times -6y+150.

$$x=-\frac{6}{5}y+30$$

Substitute -\frac{6y}{5}+30 for x in the other equation, 3x-4y=-24.

$$3\left(-\frac{6}{5}y+30\right)-4y=-24$$

Multiply 3 times -\frac{6y}{5}+30.

$$-\frac{18}{5}y+90-4y=-24$$

Add -\frac{18y}{5} to -4y.

$$-\frac{38}{5}y+90=-24$$

Subtract 90 from both sides of the equation.

$$-\frac{38}{5}y=-114$$

Divide both sides of the equation by -\frac{38}{5}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$y=15$$

Substitute 15 for y in x=-\frac{6}{5}y+30. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=-\frac{6}{5}\times 15+30$$

Multiply -\frac{6}{5} times 15.

$$x=-18+30$$

Add 30 to -18.

$$x=12$$

The system is now solved.

$$x=12,y=15$$

Steps Using Matrices

Consider the first equation. Multiply both sides of the equation by 30, the least common multiple of 6,5.

$$5x+6y=150$$

Consider the second equation. Multiply both sides of the equation by 12, the least common multiple of 4,3.

$$3x-4y=-24$$

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

$$5x+6y=150,3x-4y=-24$$

Write the equations in matrix form.

$$\left(\begin{matrix}5&6\\3&-4\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}150\\-24\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}5&6\\3&-4\end{matrix}\right).

$$inverse(\left(\begin{matrix}5&6\\3&-4\end{matrix}\right))\left(\begin{matrix}5&6\\3&-4\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}5&6\\3&-4\end{matrix}\right))\left(\begin{matrix}150\\-24\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}5&6\\3&-4\end{matrix}\right))\left(\begin{matrix}150\\-24\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}5&6\\3&-4\end{matrix}\right))\left(\begin{matrix}150\\-24\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{4}{5\left(-4\right)-6\times 3}&-\frac{6}{5\left(-4\right)-6\times 3}\\-\frac{3}{5\left(-4\right)-6\times 3}&\frac{5}{5\left(-4\right)-6\times 3}\end{matrix}\right)\left(\begin{matrix}150\\-24\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{2}{19}&\frac{3}{19}\\\frac{3}{38}&-\frac{5}{38}\end{matrix}\right)\left(\begin{matrix}150\\-24\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{2}{19}\times 150+\frac{3}{19}\left(-24\right)\\\frac{3}{38}\times 150-\frac{5}{38}\left(-24\right)\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}12\\15\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=12,y=15$$

Steps Using Elimination

Consider the first equation. Multiply both sides of the equation by 30, the least common multiple of 6,5.

$$5x+6y=150$$

Consider the second equation. Multiply both sides of the equation by 12, the least common multiple of 4,3.

$$3x-4y=-24$$

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.

$$5x+6y=150,3x-4y=-24$$

To make 5x and 3x equal, multiply all terms on each side of the first equation by 3 and all terms on each side of the second by 5.

$$3\times 5x+3\times 6y=3\times 150,5\times 3x+5\left(-4\right)y=5\left(-24\right)$$

Simplify.

$$15x+18y=450,15x-20y=-120$$

Subtract 15x-20y=-120 from 15x+18y=450 by subtracting like terms on each side of the equal sign.

$$15x-15x+18y+20y=450+120$$

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

$$18y+20y=450+120$$

Add 18y to 20y.

$$38y=450+120$$

Add 450 to 120.

$$38y=570$$

Divide both sides by 38.

$$y=15$$

Substitute 15 for y in 3x-4y=-24. Because the resulting equation contains only one variable, you can solve for x directly.

$$3x-4\times 15=-24$$

Multiply -4 times 15.

$$3x-60=-24$$

Add 60 to both sides of the equation.

$$3x=36$$

Divide both sides by 3.

$$x=12$$

The system is now solved.

$$x=12,y=15$$