Solve {l}{27quadb=3/4b+a}{11b=16a/2}

$$\left\{ \begin{array} { l } { 27 \quad b = \frac { 3 } { 4 } b + a } \\ { 11 b = \frac { 16 a } { 2 } } \end{array} \right.$$

Solve for b, a

$$b=0$$

$$a=0$$

Steps Using Substitution

Consider the first equation. Subtract \frac{3}{4}b from both sides.

$$27b-\frac{3}{4}b=a$$

Combine 27b and -\frac{3}{4}b to get \frac{105}{4}b.

$$\frac{105}{4}b=a$$

Divide both sides of the equation by \frac{105}{4}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$b=\frac{4}{105}a$$

Substitute \frac{4a}{105} for b in the other equation, 22b-16a=0.

$$22\times \frac{4}{105}a-16a=0$$

Multiply 22 times \frac{4a}{105}.

$$\frac{88}{105}a-16a=0$$

Add \frac{88a}{105} to -16a.

$$-\frac{1592}{105}a=0$$

Divide both sides of the equation by -\frac{1592}{105}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$a=0$$

Substitute 0 for a in b=\frac{4}{105}a. Because the resulting equation contains only one variable, you can solve for b directly.

$$b=0$$

The system is now solved.

$$b=0,a=0$$

Steps Using Matrices

Consider the first equation. Subtract \frac{3}{4}b from both sides.

$$27b-\frac{3}{4}b=a$$

Combine 27b and -\frac{3}{4}b to get \frac{105}{4}b.

$$\frac{105}{4}b=a$$

Subtract a from both sides.

$$\frac{105}{4}b-a=0$$

Consider the second equation. Multiply both sides of the equation by 2.

$$22b=16a$$

Subtract 16a from both sides.

$$22b-16a=0$$

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

$$\frac{105}{4}b-a=0,22b-16a=0$$

Write the equations in matrix form.

$$\left(\begin{matrix}\frac{105}{4}&-1\\22&-16\end{matrix}\right)\left(\begin{matrix}b\\a\end{matrix}\right)=\left(\begin{matrix}0\\0\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}\frac{105}{4}&-1\\22&-16\end{matrix}\right).

$$inverse(\left(\begin{matrix}\frac{105}{4}&-1\\22&-16\end{matrix}\right))\left(\begin{matrix}\frac{105}{4}&-1\\22&-16\end{matrix}\right)\left(\begin{matrix}b\\a\end{matrix}\right)=inverse(\left(\begin{matrix}\frac{105}{4}&-1\\22&-16\end{matrix}\right))\left(\begin{matrix}0\\0\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}b\\a\end{matrix}\right)=inverse(\left(\begin{matrix}\frac{105}{4}&-1\\22&-16\end{matrix}\right))\left(\begin{matrix}0\\0\end{matrix}\right)$$

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

$$\left(\begin{matrix}b\\a\end{matrix}\right)=inverse(\left(\begin{matrix}\frac{105}{4}&-1\\22&-16\end{matrix}\right))\left(\begin{matrix}0\\0\end{matrix}\right)$$

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.

$$\left(\begin{matrix}b\\a\end{matrix}\right)=\left(\begin{matrix}-\frac{16}{\frac{105}{4}\left(-16\right)-\left(-22\right)}&-\frac{-1}{\frac{105}{4}\left(-16\right)-\left(-22\right)}\\-\frac{22}{\frac{105}{4}\left(-16\right)-\left(-22\right)}&\frac{\frac{105}{4}}{\frac{105}{4}\left(-16\right)-\left(-22\right)}\end{matrix}\right)\left(\begin{matrix}0\\0\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}b\\a\end{matrix}\right)=\left(\begin{matrix}\frac{8}{199}&-\frac{1}{398}\\\frac{11}{199}&-\frac{105}{1592}\end{matrix}\right)\left(\begin{matrix}0\\0\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}b\\a\end{matrix}\right)=\left(\begin{matrix}0\\0\end{matrix}\right)$$

Extract the matrix elements b and a.

$$b=0,a=0$$

Steps Using Elimination

Consider the first equation. Subtract \frac{3}{4}b from both sides.

$$27b-\frac{3}{4}b=a$$

Combine 27b and -\frac{3}{4}b to get \frac{105}{4}b.

$$\frac{105}{4}b=a$$

Subtract a from both sides.

$$\frac{105}{4}b-a=0$$

Consider the second equation. Multiply both sides of the equation by 2.

$$22b=16a$$

Subtract 16a from both sides.

$$22b-16a=0$$

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.

$$\frac{105}{4}b-a=0,22b-16a=0$$

To make \frac{105b}{4} and 22b equal, multiply all terms on each side of the first equation by 22 and all terms on each side of the second by \frac{105}{4}.

$$22\times \frac{105}{4}b+22\left(-1\right)a=0,\frac{105}{4}\times 22b+\frac{105}{4}\left(-16\right)a=0$$

Simplify.

$$\frac{1155}{2}b-22a=0,\frac{1155}{2}b-420a=0$$

Subtract \frac{1155}{2}b-420a=0 from \frac{1155}{2}b-22a=0 by subtracting like terms on each side of the equal sign.

$$\frac{1155}{2}b-\frac{1155}{2}b-22a+420a=0$$

Add \frac{1155b}{2} to -\frac{1155b}{2}. Terms \frac{1155b}{2} and -\frac{1155b}{2} cancel out, leaving an equation with only one variable that can be solved.

$$-22a+420a=0$$

Add -22a to 420a.

$$398a=0$$

Divide both sides by 398.

$$a=0$$

Substitute 0 for a in 22b-16a=0. Because the resulting equation contains only one variable, you can solve for b directly.

$$22b=0$$

Divide both sides by 22.

$$b=0$$

The system is now solved.

$$b=0,a=0$$