Solve {l}{3x+4y=5150}{3y+4x=5000}

$$\left. \begin{array} { l } { 3 x + 4 y = 5150 } \\ { 3 y + 4 x = 5000 } \end{array} \right.$$

Solve for x, y

$$x=650$$

$$y=800$$

Steps Using Substitution

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.

$$3x+4y=5150,4x+3y=5000$$

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

$$3x+4y=5150$$

Subtract 4y from both sides of the equation.

$$3x=-4y+5150$$

Divide both sides by 3.

$$x=\frac{1}{3}\left(-4y+5150\right)$$

Multiply \frac{1}{3} times -4y+5150.

$$x=-\frac{4}{3}y+\frac{5150}{3}$$

Substitute \frac{-4y+5150}{3} for x in the other equation, 4x+3y=5000.

$$4\left(-\frac{4}{3}y+\frac{5150}{3}\right)+3y=5000$$

Multiply 4 times \frac{-4y+5150}{3}.

$$-\frac{16}{3}y+\frac{20600}{3}+3y=5000$$

Add -\frac{16y}{3} to 3y.

$$-\frac{7}{3}y+\frac{20600}{3}=5000$$

Subtract \frac{20600}{3} from both sides of the equation.

$$-\frac{7}{3}y=-\frac{5600}{3}$$

Divide both sides of the equation by -\frac{7}{3}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$y=800$$

Substitute 800 for y in x=-\frac{4}{3}y+\frac{5150}{3}. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=-\frac{4}{3}\times 800+\frac{5150}{3}$$

Multiply -\frac{4}{3} times 800.

$$x=\frac{-3200+5150}{3}$$

Add \frac{5150}{3} to -\frac{3200}{3} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=650$$

The system is now solved.

$$x=650,y=800$$

Steps Using Matrices

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

$$3x+4y=5150,4x+3y=5000$$

Write the equations in matrix form.

$$\left(\begin{matrix}3&4\\4&3\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}5150\\5000\end{matrix}\right)$$

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

$$inverse(\left(\begin{matrix}3&4\\4&3\end{matrix}\right))\left(\begin{matrix}3&4\\4&3\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}3&4\\4&3\end{matrix}\right))\left(\begin{matrix}5150\\5000\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}3&4\\4&3\end{matrix}\right))\left(\begin{matrix}5150\\5000\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}3&4\\4&3\end{matrix}\right))\left(\begin{matrix}5150\\5000\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{3}{3\times 3-4\times 4}&-\frac{4}{3\times 3-4\times 4}\\-\frac{4}{3\times 3-4\times 4}&\frac{3}{3\times 3-4\times 4}\end{matrix}\right)\left(\begin{matrix}5150\\5000\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{3}{7}&\frac{4}{7}\\\frac{4}{7}&-\frac{3}{7}\end{matrix}\right)\left(\begin{matrix}5150\\5000\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{3}{7}\times 5150+\frac{4}{7}\times 5000\\\frac{4}{7}\times 5150-\frac{3}{7}\times 5000\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}650\\800\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=650,y=800$$

Steps Using Elimination

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.

$$3x+4y=5150,4x+3y=5000$$

To make 3x 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 3.

$$4\times 3x+4\times 4y=4\times 5150,3\times 4x+3\times 3y=3\times 5000$$

Simplify.

$$12x+16y=20600,12x+9y=15000$$

Subtract 12x+9y=15000 from 12x+16y=20600 by subtracting like terms on each side of the equal sign.

$$12x-12x+16y-9y=20600-15000$$

Add 12x to -12x. Terms 12x and -12x cancel out, leaving an equation with only one variable that can be solved.

$$16y-9y=20600-15000$$

Add 16y to -9y.

$$7y=20600-15000$$

Add 20600 to -15000.

$$7y=5600$$

Divide both sides by 7.

$$y=800$$

Substitute 800 for y in 4x+3y=5000. Because the resulting equation contains only one variable, you can solve for x directly.

$$4x+3\times 800=5000$$

Multiply 3 times 800.

$$4x+2400=5000$$

Subtract 2400 from both sides of the equation.

$$4x=2600$$

Divide both sides by 4.

$$x=650$$

The system is now solved.

$$x=650,y=800$$