Solve {l}{-47+4a+(a-b)*5+6a=0}{(b-a)*5+8b+12a+15b=0}

$$\left. \begin{array} { l } { - 47 + 4 a + ( a - b ) \times 5 + 6 a = 0 } \\ { ( b - a ) \times 5 + 8 b + 12 a + 15 b = 0 } \end{array} \right.$$

Solve for a, b

$$a = \frac{188}{65} = 2\frac{58}{65} \approx 2.892307692$$

$$b=-\frac{47}{65}\approx -0.723076923$$

Steps Using Substitution

Consider the first equation. Use the distributive property to multiply a-b by 5.

$$-47+4a+5a-5b+6a=0$$

Combine 4a and 5a to get 9a.

$$-47+9a-5b+6a=0$$

Combine 9a and 6a to get 15a.

$$-47+15a-5b=0$$

Add 47 to both sides. Anything plus zero gives itself.

$$15a-5b=47$$

Consider the second equation. Use the distributive property to multiply b-a by 5.

$$5b-5a+8b+12a+15b=0$$

Combine 5b and 8b to get 13b.

$$13b-5a+12a+15b=0$$

Combine -5a and 12a to get 7a.

$$13b+7a+15b=0$$

Combine 13b and 15b to get 28b.

$$28b+7a=0$$

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.

$$15a-5b=47,7a+28b=0$$

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

$$15a-5b=47$$

Add 5b to both sides of the equation.

$$15a=5b+47$$

Divide both sides by 15.

$$a=\frac{1}{15}\left(5b+47\right)$$

Multiply \frac{1}{15} times 5b+47.

$$a=\frac{1}{3}b+\frac{47}{15}$$

Substitute \frac{b}{3}+\frac{47}{15} for a in the other equation, 7a+28b=0.

$$7\left(\frac{1}{3}b+\frac{47}{15}\right)+28b=0$$

Multiply 7 times \frac{b}{3}+\frac{47}{15}.

$$\frac{7}{3}b+\frac{329}{15}+28b=0$$

Add \frac{7b}{3} to 28b.

$$\frac{91}{3}b+\frac{329}{15}=0$$

Subtract \frac{329}{15} from both sides of the equation.

$$\frac{91}{3}b=-\frac{329}{15}$$

Divide both sides of the equation by \frac{91}{3}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$b=-\frac{47}{65}$$

Substitute -\frac{47}{65} for b in a=\frac{1}{3}b+\frac{47}{15}. Because the resulting equation contains only one variable, you can solve for a directly.

$$a=\frac{1}{3}\left(-\frac{47}{65}\right)+\frac{47}{15}$$

Multiply \frac{1}{3} times -\frac{47}{65} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$a=-\frac{47}{195}+\frac{47}{15}$$

Add \frac{47}{15} to -\frac{47}{195} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$a=\frac{188}{65}$$

The system is now solved.

$$a=\frac{188}{65},b=-\frac{47}{65}$$

Steps Using Matrices

Consider the first equation. Use the distributive property to multiply a-b by 5.

$$-47+4a+5a-5b+6a=0$$

Combine 4a and 5a to get 9a.

$$-47+9a-5b+6a=0$$

Combine 9a and 6a to get 15a.

$$-47+15a-5b=0$$

Add 47 to both sides. Anything plus zero gives itself.

$$15a-5b=47$$

Consider the second equation. Use the distributive property to multiply b-a by 5.

$$5b-5a+8b+12a+15b=0$$

Combine 5b and 8b to get 13b.

$$13b-5a+12a+15b=0$$

Combine -5a and 12a to get 7a.

$$13b+7a+15b=0$$

Combine 13b and 15b to get 28b.

$$28b+7a=0$$

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

$$15a-5b=47,7a+28b=0$$

Write the equations in matrix form.

$$\left(\begin{matrix}15&-5\\7&28\end{matrix}\right)\left(\begin{matrix}a\\b\end{matrix}\right)=\left(\begin{matrix}47\\0\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}15&-5\\7&28\end{matrix}\right).

$$inverse(\left(\begin{matrix}15&-5\\7&28\end{matrix}\right))\left(\begin{matrix}15&-5\\7&28\end{matrix}\right)\left(\begin{matrix}a\\b\end{matrix}\right)=inverse(\left(\begin{matrix}15&-5\\7&28\end{matrix}\right))\left(\begin{matrix}47\\0\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}a\\b\end{matrix}\right)=inverse(\left(\begin{matrix}15&-5\\7&28\end{matrix}\right))\left(\begin{matrix}47\\0\end{matrix}\right)$$

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

$$\left(\begin{matrix}a\\b\end{matrix}\right)=inverse(\left(\begin{matrix}15&-5\\7&28\end{matrix}\right))\left(\begin{matrix}47\\0\end{matrix}\right)$$

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.

$$\left(\begin{matrix}a\\b\end{matrix}\right)=\left(\begin{matrix}\frac{28}{15\times 28-\left(-5\times 7\right)}&-\frac{-5}{15\times 28-\left(-5\times 7\right)}\\-\frac{7}{15\times 28-\left(-5\times 7\right)}&\frac{15}{15\times 28-\left(-5\times 7\right)}\end{matrix}\right)\left(\begin{matrix}47\\0\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}a\\b\end{matrix}\right)=\left(\begin{matrix}\frac{4}{65}&\frac{1}{91}\\-\frac{1}{65}&\frac{3}{91}\end{matrix}\right)\left(\begin{matrix}47\\0\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}a\\b\end{matrix}\right)=\left(\begin{matrix}\frac{4}{65}\times 47\\-\frac{1}{65}\times 47\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}a\\b\end{matrix}\right)=\left(\begin{matrix}\frac{188}{65}\\-\frac{47}{65}\end{matrix}\right)$$

Extract the matrix elements a and b.

$$a=\frac{188}{65},b=-\frac{47}{65}$$

Steps Using Elimination

Consider the first equation. Use the distributive property to multiply a-b by 5.

$$-47+4a+5a-5b+6a=0$$

Combine 4a and 5a to get 9a.

$$-47+9a-5b+6a=0$$

Combine 9a and 6a to get 15a.

$$-47+15a-5b=0$$

Add 47 to both sides. Anything plus zero gives itself.

$$15a-5b=47$$

Consider the second equation. Use the distributive property to multiply b-a by 5.

$$5b-5a+8b+12a+15b=0$$

Combine 5b and 8b to get 13b.

$$13b-5a+12a+15b=0$$

Combine -5a and 12a to get 7a.

$$13b+7a+15b=0$$

Combine 13b and 15b to get 28b.

$$28b+7a=0$$

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.

$$15a-5b=47,7a+28b=0$$

To make 15a and 7a equal, multiply all terms on each side of the first equation by 7 and all terms on each side of the second by 15.

$$7\times 15a+7\left(-5\right)b=7\times 47,15\times 7a+15\times 28b=0$$

Simplify.

$$105a-35b=329,105a+420b=0$$

Subtract 105a+420b=0 from 105a-35b=329 by subtracting like terms on each side of the equal sign.

$$105a-105a-35b-420b=329$$

Add 105a to -105a. Terms 105a and -105a cancel out, leaving an equation with only one variable that can be solved.

$$-35b-420b=329$$

Add -35b to -420b.

$$-455b=329$$

Divide both sides by -455.

$$b=-\frac{47}{65}$$

Substitute -\frac{47}{65} for b in 7a+28b=0. Because the resulting equation contains only one variable, you can solve for a directly.

$$7a+28\left(-\frac{47}{65}\right)=0$$

Multiply 28 times -\frac{47}{65}.

$$7a-\frac{1316}{65}=0$$

Add \frac{1316}{65} to both sides of the equation.

$$7a=\frac{1316}{65}$$

Divide both sides by 7.

$$a=\frac{188}{65}$$

The system is now solved.

$$a=\frac{188}{65},b=-\frac{47}{65}$$