Solve {l}{0=16a-4b-4}{0=4a+2b-4}

$$\left\{ \begin{array} { l } { 0 = 16 a - 4 b - 4 } \\ { 0 = 4 a + 2 b - 4 } \end{array} \right.$$

Solve for a, b

$$a=\frac{1}{2}=0.5$$

$$b=1$$

Steps Using Substitution

Consider the first equation. Swap sides so that all variable terms are on the left hand side.

$$16a-4b-4=0$$

Add 4 to both sides. Anything plus zero gives itself.

$$16a-4b=4$$

Consider the second equation. Swap sides so that all variable terms are on the left hand side.

$$4a+2b-4=0$$

Add 4 to both sides. Anything plus zero gives itself.

$$4a+2b=4$$

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.

$$16a-4b=4,4a+2b=4$$

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

$$16a-4b=4$$

Add 4b to both sides of the equation.

$$16a=4b+4$$

Divide both sides by 16.

$$a=\frac{1}{16}\left(4b+4\right)$$

Multiply \frac{1}{16} times 4+4b.

$$a=\frac{1}{4}b+\frac{1}{4}$$

Substitute \frac{1+b}{4} for a in the other equation, 4a+2b=4.

$$4\left(\frac{1}{4}b+\frac{1}{4}\right)+2b=4$$

Multiply 4 times \frac{1+b}{4}.

$$b+1+2b=4$$

Add b to 2b.

$$3b+1=4$$

Subtract 1 from both sides of the equation.

$$3b=3$$

Divide both sides by 3.

$$b=1$$

Substitute 1 for b in a=\frac{1}{4}b+\frac{1}{4}. Because the resulting equation contains only one variable, you can solve for a directly.

$$a=\frac{1+1}{4}$$

Add \frac{1}{4} to \frac{1}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$a=\frac{1}{2}$$

The system is now solved.

$$a=\frac{1}{2},b=1$$

Steps Using Matrices

Consider the first equation. Swap sides so that all variable terms are on the left hand side.

$$16a-4b-4=0$$

Add 4 to both sides. Anything plus zero gives itself.

$$16a-4b=4$$

Consider the second equation. Swap sides so that all variable terms are on the left hand side.

$$4a+2b-4=0$$

Add 4 to both sides. Anything plus zero gives itself.

$$4a+2b=4$$

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

$$16a-4b=4,4a+2b=4$$

Write the equations in matrix form.

$$\left(\begin{matrix}16&-4\\4&2\end{matrix}\right)\left(\begin{matrix}a\\b\end{matrix}\right)=\left(\begin{matrix}4\\4\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}16&-4\\4&2\end{matrix}\right).

$$inverse(\left(\begin{matrix}16&-4\\4&2\end{matrix}\right))\left(\begin{matrix}16&-4\\4&2\end{matrix}\right)\left(\begin{matrix}a\\b\end{matrix}\right)=inverse(\left(\begin{matrix}16&-4\\4&2\end{matrix}\right))\left(\begin{matrix}4\\4\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}a\\b\end{matrix}\right)=inverse(\left(\begin{matrix}16&-4\\4&2\end{matrix}\right))\left(\begin{matrix}4\\4\end{matrix}\right)$$

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

$$\left(\begin{matrix}a\\b\end{matrix}\right)=inverse(\left(\begin{matrix}16&-4\\4&2\end{matrix}\right))\left(\begin{matrix}4\\4\end{matrix}\right)$$

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.

$$\left(\begin{matrix}a\\b\end{matrix}\right)=\left(\begin{matrix}\frac{2}{16\times 2-\left(-4\times 4\right)}&-\frac{-4}{16\times 2-\left(-4\times 4\right)}\\-\frac{4}{16\times 2-\left(-4\times 4\right)}&\frac{16}{16\times 2-\left(-4\times 4\right)}\end{matrix}\right)\left(\begin{matrix}4\\4\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}a\\b\end{matrix}\right)=\left(\begin{matrix}\frac{1}{24}&\frac{1}{12}\\-\frac{1}{12}&\frac{1}{3}\end{matrix}\right)\left(\begin{matrix}4\\4\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}a\\b\end{matrix}\right)=\left(\begin{matrix}\frac{1}{24}\times 4+\frac{1}{12}\times 4\\-\frac{1}{12}\times 4+\frac{1}{3}\times 4\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}a\\b\end{matrix}\right)=\left(\begin{matrix}\frac{1}{2}\\1\end{matrix}\right)$$

Extract the matrix elements a and b.

$$a=\frac{1}{2},b=1$$

Steps Using Elimination

Consider the first equation. Swap sides so that all variable terms are on the left hand side.

$$16a-4b-4=0$$

Add 4 to both sides. Anything plus zero gives itself.

$$16a-4b=4$$

Consider the second equation. Swap sides so that all variable terms are on the left hand side.

$$4a+2b-4=0$$

Add 4 to both sides. Anything plus zero gives itself.

$$4a+2b=4$$

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.

$$16a-4b=4,4a+2b=4$$

To make 16a and 4a equal, multiply all terms on each side of the first equation by 4 and all terms on each side of the second by 16.

$$4\times 16a+4\left(-4\right)b=4\times 4,16\times 4a+16\times 2b=16\times 4$$

Simplify.

$$64a-16b=16,64a+32b=64$$

Subtract 64a+32b=64 from 64a-16b=16 by subtracting like terms on each side of the equal sign.

$$64a-64a-16b-32b=16-64$$

Add 64a to -64a. Terms 64a and -64a cancel out, leaving an equation with only one variable that can be solved.

$$-16b-32b=16-64$$

Add -16b to -32b.

$$-48b=16-64$$

Add 16 to -64.

$$-48b=-48$$

Divide both sides by -48.

$$b=1$$

Substitute 1 for b in 4a+2b=4. Because the resulting equation contains only one variable, you can solve for a directly.

$$4a+2=4$$

Subtract 2 from both sides of the equation.

$$4a=2$$

Divide both sides by 4.

$$a=\frac{1}{2}$$

The system is now solved.

$$a=\frac{1}{2},b=1$$