Solve {l}{2a+7b=14}{12(2a+12b)/2=129}

$$\left. \begin{array} { l } { 2 a + 7 b = 14 } \\ { \frac { 12 ( 2 a + 12 b ) } { 2 } = 129 } \end{array} \right.$$

Solve for a, b

$$a = \frac{7}{4} = 1\frac{3}{4} = 1.75$$

$$b = \frac{3}{2} = 1\frac{1}{2} = 1.5$$

Steps Using Substitution

Consider the second equation. Multiply both sides by 2.

$$12\left(2a+12b\right)=129\times 2$$

Use the distributive property to multiply 12 by 2a+12b.

$$24a+144b=129\times 2$$

Multiply 129 and 2 to get 258.

$$24a+144b=258$$

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.

$$2a+7b=14,24a+144b=258$$

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

$$2a+7b=14$$

Subtract 7b from both sides of the equation.

$$2a=-7b+14$$

Divide both sides by 2.

$$a=\frac{1}{2}\left(-7b+14\right)$$

Multiply \frac{1}{2} times -7b+14.

$$a=-\frac{7}{2}b+7$$

Substitute -\frac{7b}{2}+7 for a in the other equation, 24a+144b=258.

$$24\left(-\frac{7}{2}b+7\right)+144b=258$$

Multiply 24 times -\frac{7b}{2}+7.

$$-84b+168+144b=258$$

Add -84b to 144b.

$$60b+168=258$$

Subtract 168 from both sides of the equation.

$$60b=90$$

Divide both sides by 60.

$$b=\frac{3}{2}$$

Substitute \frac{3}{2} for b in a=-\frac{7}{2}b+7. Because the resulting equation contains only one variable, you can solve for a directly.

$$a=-\frac{7}{2}\times \frac{3}{2}+7$$

Multiply -\frac{7}{2} times \frac{3}{2} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$a=-\frac{21}{4}+7$$

Add 7 to -\frac{21}{4}.

$$a=\frac{7}{4}$$

The system is now solved.

$$a=\frac{7}{4},b=\frac{3}{2}$$

Steps Using Matrices

Consider the second equation. Multiply both sides by 2.

$$12\left(2a+12b\right)=129\times 2$$

Use the distributive property to multiply 12 by 2a+12b.

$$24a+144b=129\times 2$$

Multiply 129 and 2 to get 258.

$$24a+144b=258$$

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

$$2a+7b=14,24a+144b=258$$

Write the equations in matrix form.

$$\left(\begin{matrix}2&7\\24&144\end{matrix}\right)\left(\begin{matrix}a\\b\end{matrix}\right)=\left(\begin{matrix}14\\258\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}2&7\\24&144\end{matrix}\right).

$$inverse(\left(\begin{matrix}2&7\\24&144\end{matrix}\right))\left(\begin{matrix}2&7\\24&144\end{matrix}\right)\left(\begin{matrix}a\\b\end{matrix}\right)=inverse(\left(\begin{matrix}2&7\\24&144\end{matrix}\right))\left(\begin{matrix}14\\258\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}a\\b\end{matrix}\right)=inverse(\left(\begin{matrix}2&7\\24&144\end{matrix}\right))\left(\begin{matrix}14\\258\end{matrix}\right)$$

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

$$\left(\begin{matrix}a\\b\end{matrix}\right)=inverse(\left(\begin{matrix}2&7\\24&144\end{matrix}\right))\left(\begin{matrix}14\\258\end{matrix}\right)$$

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.

$$\left(\begin{matrix}a\\b\end{matrix}\right)=\left(\begin{matrix}\frac{144}{2\times 144-7\times 24}&-\frac{7}{2\times 144-7\times 24}\\-\frac{24}{2\times 144-7\times 24}&\frac{2}{2\times 144-7\times 24}\end{matrix}\right)\left(\begin{matrix}14\\258\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}a\\b\end{matrix}\right)=\left(\begin{matrix}\frac{6}{5}&-\frac{7}{120}\\-\frac{1}{5}&\frac{1}{60}\end{matrix}\right)\left(\begin{matrix}14\\258\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}a\\b\end{matrix}\right)=\left(\begin{matrix}\frac{6}{5}\times 14-\frac{7}{120}\times 258\\-\frac{1}{5}\times 14+\frac{1}{60}\times 258\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}a\\b\end{matrix}\right)=\left(\begin{matrix}\frac{7}{4}\\\frac{3}{2}\end{matrix}\right)$$

Extract the matrix elements a and b.

$$a=\frac{7}{4},b=\frac{3}{2}$$

Steps Using Elimination

Consider the second equation. Multiply both sides by 2.

$$12\left(2a+12b\right)=129\times 2$$

Use the distributive property to multiply 12 by 2a+12b.

$$24a+144b=129\times 2$$

Multiply 129 and 2 to get 258.

$$24a+144b=258$$

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.

$$2a+7b=14,24a+144b=258$$

To make 2a and 24a equal, multiply all terms on each side of the first equation by 24 and all terms on each side of the second by 2.

$$24\times 2a+24\times 7b=24\times 14,2\times 24a+2\times 144b=2\times 258$$

Simplify.

$$48a+168b=336,48a+288b=516$$

Subtract 48a+288b=516 from 48a+168b=336 by subtracting like terms on each side of the equal sign.

$$48a-48a+168b-288b=336-516$$

Add 48a to -48a. Terms 48a and -48a cancel out, leaving an equation with only one variable that can be solved.

$$168b-288b=336-516$$

Add 168b to -288b.

$$-120b=336-516$$

Add 336 to -516.

$$-120b=-180$$

Divide both sides by -120.

$$b=\frac{3}{2}$$

Substitute \frac{3}{2} for b in 24a+144b=258. Because the resulting equation contains only one variable, you can solve for a directly.

$$24a+144\times \frac{3}{2}=258$$

Multiply 144 times \frac{3}{2}.

$$24a+216=258$$

Subtract 216 from both sides of the equation.

$$24a=42$$

Divide both sides by 24.

$$a=\frac{7}{4}$$

The system is now solved.

$$a=\frac{7}{4},b=\frac{3}{2}$$