Solve {l}{2a=3b}{7a+2b=200}

$$\left\{ \begin{array} { l } { 2 a = 3 b } \\ { 7 a + 2 b = 200 } \end{array} \right.$$

Solve for a, b

$$a=24$$

$$b=16$$

Steps Using Substitution

Consider the first equation. Subtract 3b from both sides.

$$2a-3b=0$$

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.

$$2a-3b=0,7a+2b=200$$

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

$$2a-3b=0$$

Add 3b to both sides of the equation.

$$2a=3b$$

Divide both sides by 2.

$$a=\frac{1}{2}\times 3b$$

Multiply \frac{1}{2} times 3b.

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

Substitute \frac{3b}{2} for a in the other equation, 7a+2b=200.

$$7\times \frac{3}{2}b+2b=200$$

Multiply 7 times \frac{3b}{2}.

$$\frac{21}{2}b+2b=200$$

Add \frac{21b}{2} to 2b.

$$\frac{25}{2}b=200$$

Divide both sides of the equation by \frac{25}{2}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$b=16$$

Substitute 16 for b in a=\frac{3}{2}b. Because the resulting equation contains only one variable, you can solve for a directly.

$$a=\frac{3}{2}\times 16$$

Multiply \frac{3}{2} times 16.

$$a=24$$

The system is now solved.

$$a=24,b=16$$

Steps Using Matrices

Consider the first equation. Subtract 3b from both sides.

$$2a-3b=0$$

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

$$2a-3b=0,7a+2b=200$$

Write the equations in matrix form.

$$\left(\begin{matrix}2&-3\\7&2\end{matrix}\right)\left(\begin{matrix}a\\b\end{matrix}\right)=\left(\begin{matrix}0\\200\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}2&-3\\7&2\end{matrix}\right).

$$inverse(\left(\begin{matrix}2&-3\\7&2\end{matrix}\right))\left(\begin{matrix}2&-3\\7&2\end{matrix}\right)\left(\begin{matrix}a\\b\end{matrix}\right)=inverse(\left(\begin{matrix}2&-3\\7&2\end{matrix}\right))\left(\begin{matrix}0\\200\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}a\\b\end{matrix}\right)=inverse(\left(\begin{matrix}2&-3\\7&2\end{matrix}\right))\left(\begin{matrix}0\\200\end{matrix}\right)$$

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

$$\left(\begin{matrix}a\\b\end{matrix}\right)=inverse(\left(\begin{matrix}2&-3\\7&2\end{matrix}\right))\left(\begin{matrix}0\\200\end{matrix}\right)$$

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.

$$\left(\begin{matrix}a\\b\end{matrix}\right)=\left(\begin{matrix}\frac{2}{2\times 2-\left(-3\times 7\right)}&-\frac{-3}{2\times 2-\left(-3\times 7\right)}\\-\frac{7}{2\times 2-\left(-3\times 7\right)}&\frac{2}{2\times 2-\left(-3\times 7\right)}\end{matrix}\right)\left(\begin{matrix}0\\200\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}a\\b\end{matrix}\right)=\left(\begin{matrix}\frac{2}{25}&\frac{3}{25}\\-\frac{7}{25}&\frac{2}{25}\end{matrix}\right)\left(\begin{matrix}0\\200\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}a\\b\end{matrix}\right)=\left(\begin{matrix}\frac{3}{25}\times 200\\\frac{2}{25}\times 200\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}a\\b\end{matrix}\right)=\left(\begin{matrix}24\\16\end{matrix}\right)$$

Extract the matrix elements a and b.

$$a=24,b=16$$

Steps Using Elimination

Consider the first equation. Subtract 3b from both sides.

$$2a-3b=0$$

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.

$$2a-3b=0,7a+2b=200$$

To make 2a 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 2.

$$7\times 2a+7\left(-3\right)b=0,2\times 7a+2\times 2b=2\times 200$$

Simplify.

$$14a-21b=0,14a+4b=400$$

Subtract 14a+4b=400 from 14a-21b=0 by subtracting like terms on each side of the equal sign.

$$14a-14a-21b-4b=-400$$

Add 14a to -14a. Terms 14a and -14a cancel out, leaving an equation with only one variable that can be solved.

$$-21b-4b=-400$$

Add -21b to -4b.

$$-25b=-400$$

Divide both sides by -25.

$$b=16$$

Substitute 16 for b in 7a+2b=200. Because the resulting equation contains only one variable, you can solve for a directly.

$$7a+2\times 16=200$$

Multiply 2 times 16.

$$7a+32=200$$

Subtract 32 from both sides of the equation.

$$7a=168$$

Divide both sides by 7.

$$a=24$$

The system is now solved.

$$a=24,b=16$$