Solve {l}{x=y-5/2}{4x+3y=65}

$$\left. \begin{array} { l } { x = \frac { y - 5 } { 2 } } \\ { 4 x + 3 y = 65 } \end{array} \right.$$

Solve for x, y

$$x=5$$

$$y=15$$

Steps Using Substitution

Consider the first equation. Divide each term of y-5 by 2 to get \frac{1}{2}y-\frac{5}{2}.

$$x=\frac{1}{2}y-\frac{5}{2}$$

Substitute \frac{-5+y}{2} for x in the other equation, 4x+3y=65.

$$4\left(\frac{1}{2}y-\frac{5}{2}\right)+3y=65$$

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

$$2y-10+3y=65$$

Add 2y to 3y.

$$5y-10=65$$

Add 10 to both sides of the equation.

$$5y=75$$

Divide both sides by 5.

$$y=15$$

Substitute 15 for y in x=\frac{1}{2}y-\frac{5}{2}. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=\frac{1}{2}\times 15-\frac{5}{2}$$

Multiply \frac{1}{2} times 15.

$$x=\frac{15-5}{2}$$

Add -\frac{5}{2} to \frac{15}{2} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=5$$

The system is now solved.

$$x=5,y=15$$

Steps Using Matrices

Consider the first equation. Divide each term of y-5 by 2 to get \frac{1}{2}y-\frac{5}{2}.

$$x=\frac{1}{2}y-\frac{5}{2}$$

Subtract \frac{1}{2}y from both sides.

$$x-\frac{1}{2}y=-\frac{5}{2}$$

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

$$x-\frac{1}{2}y=-\frac{5}{2},4x+3y=65$$

Write the equations in matrix form.

$$\left(\begin{matrix}1&-\frac{1}{2}\\4&3\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{5}{2}\\65\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}1&-\frac{1}{2}\\4&3\end{matrix}\right).

$$inverse(\left(\begin{matrix}1&-\frac{1}{2}\\4&3\end{matrix}\right))\left(\begin{matrix}1&-\frac{1}{2}\\4&3\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&-\frac{1}{2}\\4&3\end{matrix}\right))\left(\begin{matrix}-\frac{5}{2}\\65\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&-\frac{1}{2}\\4&3\end{matrix}\right))\left(\begin{matrix}-\frac{5}{2}\\65\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&-\frac{1}{2}\\4&3\end{matrix}\right))\left(\begin{matrix}-\frac{5}{2}\\65\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{3}{3-\left(-\frac{1}{2}\times 4\right)}&-\frac{-\frac{1}{2}}{3-\left(-\frac{1}{2}\times 4\right)}\\-\frac{4}{3-\left(-\frac{1}{2}\times 4\right)}&\frac{1}{3-\left(-\frac{1}{2}\times 4\right)}\end{matrix}\right)\left(\begin{matrix}-\frac{5}{2}\\65\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{3}{5}&\frac{1}{10}\\-\frac{4}{5}&\frac{1}{5}\end{matrix}\right)\left(\begin{matrix}-\frac{5}{2}\\65\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{3}{5}\left(-\frac{5}{2}\right)+\frac{1}{10}\times 65\\-\frac{4}{5}\left(-\frac{5}{2}\right)+\frac{1}{5}\times 65\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}5\\15\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=5,y=15$$

Steps Using Elimination

Consider the first equation. Divide each term of y-5 by 2 to get \frac{1}{2}y-\frac{5}{2}.

$$x=\frac{1}{2}y-\frac{5}{2}$$

Subtract \frac{1}{2}y from both sides.

$$x-\frac{1}{2}y=-\frac{5}{2}$$

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.

$$x-\frac{1}{2}y=-\frac{5}{2},4x+3y=65$$

To make x and 4x equal, multiply all terms on each side of the first equation by 4 and all terms on each side of the second by 1.

$$4x+4\left(-\frac{1}{2}\right)y=4\left(-\frac{5}{2}\right),4x+3y=65$$

Simplify.

$$4x-2y=-10,4x+3y=65$$

Subtract 4x+3y=65 from 4x-2y=-10 by subtracting like terms on each side of the equal sign.

$$4x-4x-2y-3y=-10-65$$

Add 4x to -4x. Terms 4x and -4x cancel out, leaving an equation with only one variable that can be solved.

$$-2y-3y=-10-65$$

Add -2y to -3y.

$$-5y=-10-65$$

Add -10 to -65.

$$-5y=-75$$

Divide both sides by -5.

$$y=15$$

Substitute 15 for y in 4x+3y=65. Because the resulting equation contains only one variable, you can solve for x directly.

$$4x+3\times 15=65$$

Multiply 3 times 15.

$$4x+45=65$$

Subtract 45 from both sides of the equation.

$$4x=20$$

Divide both sides by 4.

$$x=5$$

The system is now solved.

$$x=5,y=15$$