Solve {l}{x=y+1400}{0.09x+0.04y=880}

$$\left\{ \begin{array} { l } { x = y + 1400 } \\ { 0.09 x + 0.04 y = 880 } \end{array} \right.$$

Solve for x, y

$$x=7200$$

$$y=5800$$

Steps Using Substitution

Consider the first equation. Subtract y from both sides.

$$x-y=1400$$

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.

$$x-y=1400,0.09x+0.04y=880$$

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

$$x-y=1400$$

Add y to both sides of the equation.

$$x=y+1400$$

Substitute y+1400 for x in the other equation, 0.09x+0.04y=880.

$$0.09\left(y+1400\right)+0.04y=880$$

Multiply 0.09 times y+1400.

$$0.09y+126+0.04y=880$$

Add \frac{9y}{100} to \frac{y}{25}.

$$0.13y+126=880$$

Subtract 126 from both sides of the equation.

$$0.13y=754$$

Divide both sides of the equation by 0.13, which is the same as multiplying both sides by the reciprocal of the fraction.

$$y=5800$$

Substitute 5800 for y in x=y+1400. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=5800+1400$$

Add 1400 to 5800.

$$x=7200$$

The system is now solved.

$$x=7200,y=5800$$

Steps Using Matrices

Consider the first equation. Subtract y from both sides.

$$x-y=1400$$

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

$$x-y=1400,0.09x+0.04y=880$$

Write the equations in matrix form.

$$\left(\begin{matrix}1&-1\\0.09&0.04\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}1400\\880\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}1&-1\\0.09&0.04\end{matrix}\right).

$$inverse(\left(\begin{matrix}1&-1\\0.09&0.04\end{matrix}\right))\left(\begin{matrix}1&-1\\0.09&0.04\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&-1\\0.09&0.04\end{matrix}\right))\left(\begin{matrix}1400\\880\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&-1\\0.09&0.04\end{matrix}\right))\left(\begin{matrix}1400\\880\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&-1\\0.09&0.04\end{matrix}\right))\left(\begin{matrix}1400\\880\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{0.04}{0.04-\left(-0.09\right)}&-\frac{-1}{0.04-\left(-0.09\right)}\\-\frac{0.09}{0.04-\left(-0.09\right)}&\frac{1}{0.04-\left(-0.09\right)}\end{matrix}\right)\left(\begin{matrix}1400\\880\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{4}{13}&\frac{100}{13}\\-\frac{9}{13}&\frac{100}{13}\end{matrix}\right)\left(\begin{matrix}1400\\880\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{4}{13}\times 1400+\frac{100}{13}\times 880\\-\frac{9}{13}\times 1400+\frac{100}{13}\times 880\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}7200\\5800\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=7200,y=5800$$

Steps Using Elimination

Consider the first equation. Subtract y from both sides.

$$x-y=1400$$

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.

$$x-y=1400,0.09x+0.04y=880$$

To make x and \frac{9x}{100} equal, multiply all terms on each side of the first equation by 0.09 and all terms on each side of the second by 1.

$$0.09x+0.09\left(-1\right)y=0.09\times 1400,0.09x+0.04y=880$$

Simplify.

$$0.09x-0.09y=126,0.09x+0.04y=880$$

Subtract 0.09x+0.04y=880 from 0.09x-0.09y=126 by subtracting like terms on each side of the equal sign.

$$0.09x-0.09x-0.09y-0.04y=126-880$$

Add \frac{9x}{100} to -\frac{9x}{100}. Terms \frac{9x}{100} and -\frac{9x}{100} cancel out, leaving an equation with only one variable that can be solved.

$$-0.09y-0.04y=126-880$$

Add -\frac{9y}{100} to -\frac{y}{25}.

$$-0.13y=126-880$$

Add 126 to -880.

$$-0.13y=-754$$

Divide both sides of the equation by -0.13, which is the same as multiplying both sides by the reciprocal of the fraction.

$$y=5800$$

Substitute 5800 for y in 0.09x+0.04y=880. Because the resulting equation contains only one variable, you can solve for x directly.

$$0.09x+0.04\times 5800=880$$

Multiply 0.04 times 5800.

$$0.09x+232=880$$

Subtract 232 from both sides of the equation.

$$0.09x=648$$

Divide both sides of the equation by 0.09, which is the same as multiplying both sides by the reciprocal of the fraction.

$$x=7200$$

The system is now solved.

$$x=7200,y=5800$$