Solve {l}{x+y=42000}{a*12x+0.08y=(42000*1.1)+80}

$$\left. \begin{array} { l } { x + y = 42000 } \\ { a \cdot 12 x + 0.08 y = ( 42000 \times 1.1 ) + 80 } \end{array} \right.$$

Solve for x, y

$$x=\frac{536500}{150a-1}$$

$$y=-\frac{500\left(1157-12600a\right)}{150a-1}$$

$$a\neq \frac{1}{150}$$

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.

$$x+y=42000,12ax+0.08y=46280$$

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

$$x+y=42000$$

Subtract y from both sides of the equation.

$$x=-y+42000$$

Substitute -y+42000 for x in the other equation, 12ax+0.08y=46280.

$$12a\left(-y+42000\right)+0.08y=46280$$

Multiply 12a times -y+42000.

$$\left(-12a\right)y+504000a+0.08y=46280$$

Add -12ay to \frac{2y}{25}.

$$\left(0.08-12a\right)y+504000a=46280$$

Subtract 504000a from both sides of the equation.

$$\left(0.08-12a\right)y=46280-504000a$$

Divide both sides by -12a+0.08.

$$y=\frac{40\left(1157-12600a\right)}{0.08-12a}$$

Substitute \frac{40\left(1157-12600a\right)}{-12a+0.08} for y in x=-y+42000. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=-\frac{40\left(1157-12600a\right)}{0.08-12a}+42000$$

Add 42000 to -\frac{40\left(1157-12600a\right)}{-12a+0.08}.

$$x=-\frac{42920}{0.08-12a}$$

The system is now solved.

$$x=-\frac{42920}{0.08-12a},y=\frac{40\left(1157-12600a\right)}{0.08-12a}$$

Steps Using Matrices

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

$$x+y=42000,12ax+0.08y=46280$$

Write the equations in matrix form.

$$\left(\begin{matrix}1&1\\12a&0.08\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}42000\\46280\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}1&1\\12a&0.08\end{matrix}\right).

$$inverse(\left(\begin{matrix}1&1\\12a&0.08\end{matrix}\right))\left(\begin{matrix}1&1\\12a&0.08\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\12a&0.08\end{matrix}\right))\left(\begin{matrix}42000\\46280\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\12a&0.08\end{matrix}\right))\left(\begin{matrix}42000\\46280\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\12a&0.08\end{matrix}\right))\left(\begin{matrix}42000\\46280\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{0.08}{0.08-12a}&-\frac{1}{0.08-12a}\\-\frac{12a}{0.08-12a}&\frac{1}{0.08-12a}\end{matrix}\right)\left(\begin{matrix}42000\\46280\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{2}{25\left(0.08-12a\right)}&-\frac{1}{0.08-12a}\\-\frac{12a}{0.08-12a}&\frac{1}{0.08-12a}\end{matrix}\right)\left(\begin{matrix}42000\\46280\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{2}{25\left(0.08-12a\right)}\times 42000+\left(-\frac{1}{0.08-12a}\right)\times 46280\\\left(-\frac{12a}{0.08-12a}\right)\times 42000+\frac{1}{0.08-12a}\times 46280\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{42920}{0.08-12a}\\\frac{480\left(12600a-1157\right)}{144a-0.96}\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=-\frac{42920}{0.08-12a},y=\frac{480\left(12600a-1157\right)}{144a-0.96}$$

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.

$$x+y=42000,12ax+0.08y=46280$$

To make x and 12ax equal, multiply all terms on each side of the first equation by 12a and all terms on each side of the second by 1.

$$12ax+12ay=12a\times 42000,12ax+0.08y=46280$$

Simplify.

$$12ax+12ay=504000a,12ax+0.08y=46280$$

Subtract 12ax+0.08y=46280 from 12ax+12ay=504000a by subtracting like terms on each side of the equal sign.

$$12ax+\left(-12a\right)x+12ay-0.08y=504000a-46280$$

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

$$12ay-0.08y=504000a-46280$$

Add 12ay to -\frac{2y}{25}.

$$\left(12a-0.08\right)y=504000a-46280$$

Divide both sides by 12a-0.08.

$$y=\frac{40\left(12600a-1157\right)}{12a-0.08}$$

Substitute \frac{40\left(12600a-1157\right)}{12a-0.08} for y in 12ax+0.08y=46280. Because the resulting equation contains only one variable, you can solve for x directly.

$$12ax+0.08\times \frac{40\left(12600a-1157\right)}{12a-0.08}=46280$$

Multiply 0.08 times \frac{40\left(12600a-1157\right)}{12a-0.08}.

$$12ax+\frac{16\left(12600a-1157\right)}{5\left(12a-0.08\right)}=46280$$

Subtract \frac{16\left(12600a-1157\right)}{5\left(12a-0.08\right)} from both sides of the equation.

$$12ax=\frac{515040a}{12a-0.08}$$

Divide both sides by 12a.

$$x=\frac{42920}{12a-0.08}$$

The system is now solved.

$$x=\frac{42920}{12a-0.08},y=\frac{40\left(12600a-1157\right)}{12a-0.08}$$