Solve {l}{3-a/300+-5b-a/10-a-160/20=0}{b=a-160/20}

$$\left. \begin{array} { l } { 3 - \frac { a } { 300 } + \frac { - 5 b - a } { 10 } - \frac { a - 160 } { 20 } = 0 } \\ { b = \frac { a - 160 } { 20 } } \end{array} \right.$$

Solve for a, b

$$a = \frac{9000}{107} = 84\frac{12}{107} \approx 84.112149533$$

$$b = -\frac{406}{107} = -3\frac{85}{107} \approx -3.794392523$$

Steps Using Substitution

Consider the first equation. Multiply both sides of the equation by 300, the least common multiple of 300,10,20.

$$900-a+30\left(-5b-a\right)-15\left(a-160\right)=0$$

Use the distributive property to multiply 30 by -5b-a.

$$900-a-150b-30a-15\left(a-160\right)=0$$

Use the distributive property to multiply -15 by a-160.

$$900-a-150b-30a-15a+2400=0$$

Combine -30a and -15a to get -45a.

$$900-a-150b-45a+2400=0$$

Add 900 and 2400 to get 3300.

$$3300-a-150b-45a=0$$

Subtract 3300 from both sides. Anything subtracted from zero gives its negation.

$$-a-150b-45a=-3300$$

Combine -a and -45a to get -46a.

$$-46a-150b=-3300$$

Consider the second equation. Divide each term of a-160 by 20 to get \frac{1}{20}a-8.

$$b=\frac{1}{20}a-8$$

Subtract \frac{1}{20}a from both sides.

$$b-\frac{1}{20}a=-8$$

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.

$$-46a-150b=-3300,-\frac{1}{20}a+b=-8$$

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

$$-46a-150b=-3300$$

Add 150b to both sides of the equation.

$$-46a=150b-3300$$

Divide both sides by -46.

$$a=-\frac{1}{46}\left(150b-3300\right)$$

Multiply -\frac{1}{46} times -3300+150b.

$$a=-\frac{75}{23}b+\frac{1650}{23}$$

Substitute \frac{-75b+1650}{23} for a in the other equation, -\frac{1}{20}a+b=-8.

$$-\frac{1}{20}\left(-\frac{75}{23}b+\frac{1650}{23}\right)+b=-8$$

Multiply -\frac{1}{20} times \frac{-75b+1650}{23}.

$$\frac{15}{92}b-\frac{165}{46}+b=-8$$

Add \frac{15b}{92} to b.

$$\frac{107}{92}b-\frac{165}{46}=-8$$

Add \frac{165}{46} to both sides of the equation.

$$\frac{107}{92}b=-\frac{203}{46}$$

Divide both sides of the equation by \frac{107}{92}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$b=-\frac{406}{107}$$

Substitute -\frac{406}{107} for b in a=-\frac{75}{23}b+\frac{1650}{23}. Because the resulting equation contains only one variable, you can solve for a directly.

$$a=-\frac{75}{23}\left(-\frac{406}{107}\right)+\frac{1650}{23}$$

Multiply -\frac{75}{23} times -\frac{406}{107} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$a=\frac{30450}{2461}+\frac{1650}{23}$$

Add \frac{1650}{23} to \frac{30450}{2461} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$a=\frac{9000}{107}$$

The system is now solved.

$$a=\frac{9000}{107},b=-\frac{406}{107}$$

Steps Using Matrices

Consider the first equation. Multiply both sides of the equation by 300, the least common multiple of 300,10,20.

$$900-a+30\left(-5b-a\right)-15\left(a-160\right)=0$$

Use the distributive property to multiply 30 by -5b-a.

$$900-a-150b-30a-15\left(a-160\right)=0$$

Use the distributive property to multiply -15 by a-160.

$$900-a-150b-30a-15a+2400=0$$

Combine -30a and -15a to get -45a.

$$900-a-150b-45a+2400=0$$

Add 900 and 2400 to get 3300.

$$3300-a-150b-45a=0$$

Subtract 3300 from both sides. Anything subtracted from zero gives its negation.

$$-a-150b-45a=-3300$$

Combine -a and -45a to get -46a.

$$-46a-150b=-3300$$

Consider the second equation. Divide each term of a-160 by 20 to get \frac{1}{20}a-8.

$$b=\frac{1}{20}a-8$$

Subtract \frac{1}{20}a from both sides.

$$b-\frac{1}{20}a=-8$$

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

$$-46a-150b=-3300,-\frac{1}{20}a+b=-8$$

Write the equations in matrix form.

$$\left(\begin{matrix}-46&-150\\-\frac{1}{20}&1\end{matrix}\right)\left(\begin{matrix}a\\b\end{matrix}\right)=\left(\begin{matrix}-3300\\-8\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}-46&-150\\-\frac{1}{20}&1\end{matrix}\right).

$$inverse(\left(\begin{matrix}-46&-150\\-\frac{1}{20}&1\end{matrix}\right))\left(\begin{matrix}-46&-150\\-\frac{1}{20}&1\end{matrix}\right)\left(\begin{matrix}a\\b\end{matrix}\right)=inverse(\left(\begin{matrix}-46&-150\\-\frac{1}{20}&1\end{matrix}\right))\left(\begin{matrix}-3300\\-8\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}a\\b\end{matrix}\right)=inverse(\left(\begin{matrix}-46&-150\\-\frac{1}{20}&1\end{matrix}\right))\left(\begin{matrix}-3300\\-8\end{matrix}\right)$$

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

$$\left(\begin{matrix}a\\b\end{matrix}\right)=inverse(\left(\begin{matrix}-46&-150\\-\frac{1}{20}&1\end{matrix}\right))\left(\begin{matrix}-3300\\-8\end{matrix}\right)$$

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.

$$\left(\begin{matrix}a\\b\end{matrix}\right)=\left(\begin{matrix}\frac{1}{-46-\left(-150\left(-\frac{1}{20}\right)\right)}&-\frac{-150}{-46-\left(-150\left(-\frac{1}{20}\right)\right)}\\-\frac{-\frac{1}{20}}{-46-\left(-150\left(-\frac{1}{20}\right)\right)}&-\frac{46}{-46-\left(-150\left(-\frac{1}{20}\right)\right)}\end{matrix}\right)\left(\begin{matrix}-3300\\-8\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}a\\b\end{matrix}\right)=\left(\begin{matrix}-\frac{2}{107}&-\frac{300}{107}\\-\frac{1}{1070}&\frac{92}{107}\end{matrix}\right)\left(\begin{matrix}-3300\\-8\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}a\\b\end{matrix}\right)=\left(\begin{matrix}-\frac{2}{107}\left(-3300\right)-\frac{300}{107}\left(-8\right)\\-\frac{1}{1070}\left(-3300\right)+\frac{92}{107}\left(-8\right)\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}a\\b\end{matrix}\right)=\left(\begin{matrix}\frac{9000}{107}\\-\frac{406}{107}\end{matrix}\right)$$

Extract the matrix elements a and b.

$$a=\frac{9000}{107},b=-\frac{406}{107}$$

Steps Using Elimination

Consider the first equation. Multiply both sides of the equation by 300, the least common multiple of 300,10,20.

$$900-a+30\left(-5b-a\right)-15\left(a-160\right)=0$$

Use the distributive property to multiply 30 by -5b-a.

$$900-a-150b-30a-15\left(a-160\right)=0$$

Use the distributive property to multiply -15 by a-160.

$$900-a-150b-30a-15a+2400=0$$

Combine -30a and -15a to get -45a.

$$900-a-150b-45a+2400=0$$

Add 900 and 2400 to get 3300.

$$3300-a-150b-45a=0$$

Subtract 3300 from both sides. Anything subtracted from zero gives its negation.

$$-a-150b-45a=-3300$$

Combine -a and -45a to get -46a.

$$-46a-150b=-3300$$

Consider the second equation. Divide each term of a-160 by 20 to get \frac{1}{20}a-8.

$$b=\frac{1}{20}a-8$$

Subtract \frac{1}{20}a from both sides.

$$b-\frac{1}{20}a=-8$$

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.

$$-46a-150b=-3300,-\frac{1}{20}a+b=-8$$

To make -46a and -\frac{a}{20} equal, multiply all terms on each side of the first equation by -\frac{1}{20} and all terms on each side of the second by -46.

$$-\frac{1}{20}\left(-46\right)a-\frac{1}{20}\left(-150\right)b=-\frac{1}{20}\left(-3300\right),-46\left(-\frac{1}{20}\right)a-46b=-46\left(-8\right)$$

Simplify.

$$\frac{23}{10}a+\frac{15}{2}b=165,\frac{23}{10}a-46b=368$$

Subtract \frac{23}{10}a-46b=368 from \frac{23}{10}a+\frac{15}{2}b=165 by subtracting like terms on each side of the equal sign.

$$\frac{23}{10}a-\frac{23}{10}a+\frac{15}{2}b+46b=165-368$$

Add \frac{23a}{10} to -\frac{23a}{10}. Terms \frac{23a}{10} and -\frac{23a}{10} cancel out, leaving an equation with only one variable that can be solved.

$$\frac{15}{2}b+46b=165-368$$

Add \frac{15b}{2} to 46b.

$$\frac{107}{2}b=165-368$$

Add 165 to -368.

$$\frac{107}{2}b=-203$$

Divide both sides of the equation by \frac{107}{2}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$b=-\frac{406}{107}$$

Substitute -\frac{406}{107} for b in -\frac{1}{20}a+b=-8. Because the resulting equation contains only one variable, you can solve for a directly.

$$-\frac{1}{20}a-\frac{406}{107}=-8$$

Add \frac{406}{107} to both sides of the equation.

$$-\frac{1}{20}a=-\frac{450}{107}$$

Multiply both sides by -20.

$$a=\frac{9000}{107}$$

The system is now solved.

$$a=\frac{9000}{107},b=-\frac{406}{107}$$