Solve {l}{a-b+5=0}{25a+5b+5=0}

$$\left\{ \begin{array} { l } { a - b + 5 = 0 } \\ { 25 a + 5 b + 5 = 0 } \end{array} \right.$$

Solve for a, b

$$a=-1$$

$$b=4$$

Steps Using Substitution

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.

$$a-b+5=0,25a+5b+5=0$$

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

$$a-b+5=0$$

Subtract 5 from both sides of the equation.

$$a-b=-5$$

Add b to both sides of the equation.

$$a=b-5$$

Substitute b-5 for a in the other equation, 25a+5b+5=0.

$$25\left(b-5\right)+5b+5=0$$

Multiply 25 times b-5.

$$25b-125+5b+5=0$$

Add 25b to 5b.

$$30b-125+5=0$$

Add -125 to 5.

$$30b-120=0$$

Add 120 to both sides of the equation.

$$30b=120$$

Divide both sides by 30.

$$b=4$$

Substitute 4 for b in a=b-5. Because the resulting equation contains only one variable, you can solve for a directly.

$$a=4-5$$

Add -5 to 4.

$$a=-1$$

The system is now solved.

$$a=-1,b=4$$

Steps Using Matrices

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

$$a-b+5=0,25a+5b+5=0$$

Write the equations in matrix form.

$$\left(\begin{matrix}1&-1\\25&5\end{matrix}\right)\left(\begin{matrix}a\\b\end{matrix}\right)=\left(\begin{matrix}-5\\-5\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}1&-1\\25&5\end{matrix}\right).

$$inverse(\left(\begin{matrix}1&-1\\25&5\end{matrix}\right))\left(\begin{matrix}1&-1\\25&5\end{matrix}\right)\left(\begin{matrix}a\\b\end{matrix}\right)=inverse(\left(\begin{matrix}1&-1\\25&5\end{matrix}\right))\left(\begin{matrix}-5\\-5\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}a\\b\end{matrix}\right)=inverse(\left(\begin{matrix}1&-1\\25&5\end{matrix}\right))\left(\begin{matrix}-5\\-5\end{matrix}\right)$$

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

$$\left(\begin{matrix}a\\b\end{matrix}\right)=inverse(\left(\begin{matrix}1&-1\\25&5\end{matrix}\right))\left(\begin{matrix}-5\\-5\end{matrix}\right)$$

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.

$$\left(\begin{matrix}a\\b\end{matrix}\right)=\left(\begin{matrix}\frac{5}{5-\left(-25\right)}&-\frac{-1}{5-\left(-25\right)}\\-\frac{25}{5-\left(-25\right)}&\frac{1}{5-\left(-25\right)}\end{matrix}\right)\left(\begin{matrix}-5\\-5\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}a\\b\end{matrix}\right)=\left(\begin{matrix}\frac{1}{6}&\frac{1}{30}\\-\frac{5}{6}&\frac{1}{30}\end{matrix}\right)\left(\begin{matrix}-5\\-5\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}a\\b\end{matrix}\right)=\left(\begin{matrix}\frac{1}{6}\left(-5\right)+\frac{1}{30}\left(-5\right)\\-\frac{5}{6}\left(-5\right)+\frac{1}{30}\left(-5\right)\end{matrix}\right)$$

Do the arithmetic.

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

Extract the matrix elements a and b.

$$a=-1,b=4$$

Steps Using Elimination

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.

$$a-b+5=0,25a+5b+5=0$$

To make a and 25a equal, multiply all terms on each side of the first equation by 25 and all terms on each side of the second by 1.

$$25a+25\left(-1\right)b+25\times 5=0,25a+5b+5=0$$

Simplify.

$$25a-25b+125=0,25a+5b+5=0$$

Subtract 25a+5b+5=0 from 25a-25b+125=0 by subtracting like terms on each side of the equal sign.

$$25a-25a-25b-5b+125-5=0$$

Add 25a to -25a. Terms 25a and -25a cancel out, leaving an equation with only one variable that can be solved.

$$-25b-5b+125-5=0$$

Add -25b to -5b.

$$-30b+125-5=0$$

Add 125 to -5.

$$-30b+120=0$$

Subtract 120 from both sides of the equation.

$$-30b=-120$$

Divide both sides by -30.

$$b=4$$

Substitute 4 for b in 25a+5b+5=0. Because the resulting equation contains only one variable, you can solve for a directly.

$$25a+5\times 4+5=0$$

Multiply 5 times 4.

$$25a+20+5=0$$

Add 20 to 5.

$$25a+25=0$$

Subtract 25 from both sides of the equation.

$$25a=-25$$

Divide both sides by 25.

$$a=-1$$

The system is now solved.

$$a=-1,b=4$$