Solve {l}{a+19=15b}{a+19=1/2(b+19)+8,5}

$$\left\{ \begin{array} { l } { a + 19 = 15 b } \\ { a + 19 = \frac { 1 } { 2 } ( b + 19 ) + 8,5 } \end{array} \right.$$

Solve for a, b

$$a=-\frac{11}{29}\approx -0,379310345$$

$$b=\frac{36}{29}\approx 1,24137931$$

Steps Using Substitution

Consider the first equation. Subtract 15b from both sides.

$$a+19-15b=0$$

Subtract 19 from both sides. Anything subtracted from zero gives its negation.

$$a-15b=-19$$

Consider the second equation. Use the distributive property to multiply \frac{1}{2} by b+19.

$$a+19=\frac{1}{2}b+\frac{19}{2}+8,5$$

Add \frac{19}{2} and 8,5 to get 18.

$$a+19=\frac{1}{2}b+18$$

Subtract \frac{1}{2}b from both sides.

$$a+19-\frac{1}{2}b=18$$

Subtract 19 from both sides.

$$a-\frac{1}{2}b=18-19$$

Subtract 19 from 18 to get -1.

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

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.

$$a-15b=-19;a-\frac{1}{2}b=-1$$

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

$$a-15b=-19$$

Add 15b to both sides of the equation.

$$a=15b-19$$

Substitute 15b-19 for a in the other equation, a-\frac{1}{2}b=-1.

$$15b-19-\frac{1}{2}b=-1$$

Add 15b to -\frac{b}{2}.

$$\frac{29}{2}b-19=-1$$

Add 19 to both sides of the equation.

$$\frac{29}{2}b=18$$

Divide both sides of the equation by \frac{29}{2}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$b=\frac{36}{29}$$

Substitute \frac{36}{29} for b in a=15b-19. Because the resulting equation contains only one variable, you can solve for a directly.

$$a=15\times \frac{36}{29}-19$$

Multiply 15 times \frac{36}{29}.

$$a=\frac{540}{29}-19$$

Add -19 to \frac{540}{29}.

$$a=-\frac{11}{29}$$

The system is now solved.

$$a=-\frac{11}{29};b=\frac{36}{29}$$

Steps Using Matrices

Consider the first equation. Subtract 15b from both sides.

$$a+19-15b=0$$

Subtract 19 from both sides. Anything subtracted from zero gives its negation.

$$a-15b=-19$$

Consider the second equation. Use the distributive property to multiply \frac{1}{2} by b+19.

$$a+19=\frac{1}{2}b+\frac{19}{2}+8,5$$

Add \frac{19}{2} and 8,5 to get 18.

$$a+19=\frac{1}{2}b+18$$

Subtract \frac{1}{2}b from both sides.

$$a+19-\frac{1}{2}b=18$$

Subtract 19 from both sides.

$$a-\frac{1}{2}b=18-19$$

Subtract 19 from 18 to get -1.

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

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

$$a-15b=-19;a-\frac{1}{2}b=-1$$

Write the equations in matrix form.

$$\left(\begin{matrix}1&-15\\1&-\frac{1}{2}\end{matrix}\right)\left(\begin{matrix}a\\b\end{matrix}\right)=\left(\begin{matrix}-19\\-1\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}1&-15\\1&-\frac{1}{2}\end{matrix}\right).

$$inverse(\left(\begin{matrix}1&-15\\1&-\frac{1}{2}\end{matrix}\right))\left(\begin{matrix}1&-15\\1&-\frac{1}{2}\end{matrix}\right)\left(\begin{matrix}a\\b\end{matrix}\right)=inverse(\left(\begin{matrix}1&-15\\1&-\frac{1}{2}\end{matrix}\right))\left(\begin{matrix}-19\\-1\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}a\\b\end{matrix}\right)=inverse(\left(\begin{matrix}1&-15\\1&-\frac{1}{2}\end{matrix}\right))\left(\begin{matrix}-19\\-1\end{matrix}\right)$$

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

$$\left(\begin{matrix}a\\b\end{matrix}\right)=inverse(\left(\begin{matrix}1&-15\\1&-\frac{1}{2}\end{matrix}\right))\left(\begin{matrix}-19\\-1\end{matrix}\right)$$

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.

$$\left(\begin{matrix}a\\b\end{matrix}\right)=\left(\begin{matrix}-\frac{\frac{1}{2}}{-\frac{1}{2}-\left(-15\right)}&-\frac{-15}{-\frac{1}{2}-\left(-15\right)}\\-\frac{1}{-\frac{1}{2}-\left(-15\right)}&\frac{1}{-\frac{1}{2}-\left(-15\right)}\end{matrix}\right)\left(\begin{matrix}-19\\-1\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}a\\b\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{29}&\frac{30}{29}\\-\frac{2}{29}&\frac{2}{29}\end{matrix}\right)\left(\begin{matrix}-19\\-1\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}a\\b\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{29}\left(-19\right)+\frac{30}{29}\left(-1\right)\\-\frac{2}{29}\left(-19\right)+\frac{2}{29}\left(-1\right)\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}a\\b\end{matrix}\right)=\left(\begin{matrix}-\frac{11}{29}\\\frac{36}{29}\end{matrix}\right)$$

Extract the matrix elements a and b.

$$a=-\frac{11}{29};b=\frac{36}{29}$$

Steps Using Elimination

Consider the first equation. Subtract 15b from both sides.

$$a+19-15b=0$$

Subtract 19 from both sides. Anything subtracted from zero gives its negation.

$$a-15b=-19$$

Consider the second equation. Use the distributive property to multiply \frac{1}{2} by b+19.

$$a+19=\frac{1}{2}b+\frac{19}{2}+8,5$$

Add \frac{19}{2} and 8,5 to get 18.

$$a+19=\frac{1}{2}b+18$$

Subtract \frac{1}{2}b from both sides.

$$a+19-\frac{1}{2}b=18$$

Subtract 19 from both sides.

$$a-\frac{1}{2}b=18-19$$

Subtract 19 from 18 to get -1.

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

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.

$$a-15b=-19;a-\frac{1}{2}b=-1$$

Subtract a-\frac{1}{2}b=-1 from a-15b=-19 by subtracting like terms on each side of the equal sign.

$$a-a-15b+\frac{1}{2}b=-19+1$$

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

$$-15b+\frac{1}{2}b=-19+1$$

Add -15b to \frac{b}{2}.

$$-\frac{29}{2}b=-19+1$$

Add -19 to 1.

$$-\frac{29}{2}b=-18$$

Divide both sides of the equation by -\frac{29}{2}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$b=\frac{36}{29}$$

Substitute \frac{36}{29} for b in a-\frac{1}{2}b=-1. Because the resulting equation contains only one variable, you can solve for a directly.

$$a-\frac{1}{2}\times \frac{36}{29}=-1$$

Multiply -\frac{1}{2} times \frac{36}{29} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$a-\frac{18}{29}=-1$$

Add \frac{18}{29} to both sides of the equation.

$$a=-\frac{11}{29}$$

The system is now solved.

$$a=-\frac{11}{29};b=\frac{36}{29}$$