Solve {l}{(a-35/3)60=20x+260}{(a+15)12=20x+180}

$$\left. \begin{array} { l } { ( a - \frac { 35 } { 3 } ) 60 = 20 x + 260 } \\ { ( a + 15 ) 12 = 20 x + 180 } \end{array} \right.$$

Solve for a, x

$$x=12$$

$$a=20$$

Steps Using Substitution

Consider the first equation. Use the distributive property to multiply a-\frac{35}{3} by 60.

$$60a-700=20x+260$$

Subtract 20x from both sides.

$$60a-700-20x=260$$

Add 700 to both sides.

$$60a-20x=260+700$$

Add 260 and 700 to get 960.

$$60a-20x=960$$

Consider the second equation. Use the distributive property to multiply a+15 by 12.

$$12a+180=20x+180$$

Subtract 20x from both sides.

$$12a+180-20x=180$$

Subtract 180 from both sides.

$$12a-20x=180-180$$

Subtract 180 from 180 to get 0.

$$12a-20x=0$$

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.

$$60a-20x=960,12a-20x=0$$

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

$$60a-20x=960$$

Add 20x to both sides of the equation.

$$60a=20x+960$$

Divide both sides by 60.

$$a=\frac{1}{60}\left(20x+960\right)$$

Multiply \frac{1}{60} times 960+20x.

$$a=\frac{1}{3}x+16$$

Substitute \frac{x}{3}+16 for a in the other equation, 12a-20x=0.

$$12\left(\frac{1}{3}x+16\right)-20x=0$$

Multiply 12 times \frac{x}{3}+16.

$$4x+192-20x=0$$

Add 4x to -20x.

$$-16x+192=0$$

Subtract 192 from both sides of the equation.

$$-16x=-192$$

Divide both sides by -16.

$$x=12$$

Substitute 12 for x in a=\frac{1}{3}x+16. Because the resulting equation contains only one variable, you can solve for a directly.

$$a=\frac{1}{3}\times 12+16$$

Multiply \frac{1}{3} times 12.

$$a=4+16$$

Add 16 to 4.

$$a=20$$

The system is now solved.

$$a=20,x=12$$

Steps Using Matrices

Consider the first equation. Use the distributive property to multiply a-\frac{35}{3} by 60.

$$60a-700=20x+260$$

Subtract 20x from both sides.

$$60a-700-20x=260$$

Add 700 to both sides.

$$60a-20x=260+700$$

Add 260 and 700 to get 960.

$$60a-20x=960$$

Consider the second equation. Use the distributive property to multiply a+15 by 12.

$$12a+180=20x+180$$

Subtract 20x from both sides.

$$12a+180-20x=180$$

Subtract 180 from both sides.

$$12a-20x=180-180$$

Subtract 180 from 180 to get 0.

$$12a-20x=0$$

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

$$60a-20x=960,12a-20x=0$$

Write the equations in matrix form.

$$\left(\begin{matrix}60&-20\\12&-20\end{matrix}\right)\left(\begin{matrix}a\\x\end{matrix}\right)=\left(\begin{matrix}960\\0\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}60&-20\\12&-20\end{matrix}\right).

$$inverse(\left(\begin{matrix}60&-20\\12&-20\end{matrix}\right))\left(\begin{matrix}60&-20\\12&-20\end{matrix}\right)\left(\begin{matrix}a\\x\end{matrix}\right)=inverse(\left(\begin{matrix}60&-20\\12&-20\end{matrix}\right))\left(\begin{matrix}960\\0\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}a\\x\end{matrix}\right)=inverse(\left(\begin{matrix}60&-20\\12&-20\end{matrix}\right))\left(\begin{matrix}960\\0\end{matrix}\right)$$

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

$$\left(\begin{matrix}a\\x\end{matrix}\right)=inverse(\left(\begin{matrix}60&-20\\12&-20\end{matrix}\right))\left(\begin{matrix}960\\0\end{matrix}\right)$$

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.

$$\left(\begin{matrix}a\\x\end{matrix}\right)=\left(\begin{matrix}-\frac{20}{60\left(-20\right)-\left(-20\times 12\right)}&-\frac{-20}{60\left(-20\right)-\left(-20\times 12\right)}\\-\frac{12}{60\left(-20\right)-\left(-20\times 12\right)}&\frac{60}{60\left(-20\right)-\left(-20\times 12\right)}\end{matrix}\right)\left(\begin{matrix}960\\0\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}a\\x\end{matrix}\right)=\left(\begin{matrix}\frac{1}{48}&-\frac{1}{48}\\\frac{1}{80}&-\frac{1}{16}\end{matrix}\right)\left(\begin{matrix}960\\0\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}a\\x\end{matrix}\right)=\left(\begin{matrix}\frac{1}{48}\times 960\\\frac{1}{80}\times 960\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}a\\x\end{matrix}\right)=\left(\begin{matrix}20\\12\end{matrix}\right)$$

Extract the matrix elements a and x.

$$a=20,x=12$$

Steps Using Elimination

Consider the first equation. Use the distributive property to multiply a-\frac{35}{3} by 60.

$$60a-700=20x+260$$

Subtract 20x from both sides.

$$60a-700-20x=260$$

Add 700 to both sides.

$$60a-20x=260+700$$

Add 260 and 700 to get 960.

$$60a-20x=960$$

Consider the second equation. Use the distributive property to multiply a+15 by 12.

$$12a+180=20x+180$$

Subtract 20x from both sides.

$$12a+180-20x=180$$

Subtract 180 from both sides.

$$12a-20x=180-180$$

Subtract 180 from 180 to get 0.

$$12a-20x=0$$

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.

$$60a-20x=960,12a-20x=0$$

Subtract 12a-20x=0 from 60a-20x=960 by subtracting like terms on each side of the equal sign.

$$60a-12a-20x+20x=960$$

Add -20x to 20x. Terms -20x and 20x cancel out, leaving an equation with only one variable that can be solved.

$$60a-12a=960$$

Add 60a to -12a.

$$48a=960$$

Divide both sides by 48.

$$a=20$$

Substitute 20 for a in 12a-20x=0. Because the resulting equation contains only one variable, you can solve for x directly.

$$12\times 20-20x=0$$

Multiply 12 times 20.

$$240-20x=0$$

Subtract 240 from both sides of the equation.

$$-20x=-240$$

Divide both sides by -20.

$$x=12$$

The system is now solved.

$$a=20,x=12$$