Solve {l}{a-b=-2}{3a+4b=2429}

$$\left. \begin{array} { l } { a - b = - 2 } \\ { 3 a + 4 b = 2429 } \end{array} \right.$$

Solve for a, b

$$a = \frac{2421}{7} = 345\frac{6}{7} \approx 345.857142857$$

$$b = \frac{2435}{7} = 347\frac{6}{7} \approx 347.857142857$$

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.

$$a-b=-2,3a+4b=2429$$

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

$$a-b=-2$$

Add b to both sides of the equation.

$$a=b-2$$

Substitute b-2 for a in the other equation, 3a+4b=2429.

$$3\left(b-2\right)+4b=2429$$

Multiply 3 times b-2.

$$3b-6+4b=2429$$

Add 3b to 4b.

$$7b-6=2429$$

Add 6 to both sides of the equation.

$$7b=2435$$

Divide both sides by 7.

$$b=\frac{2435}{7}$$

Substitute \frac{2435}{7} for b in a=b-2. Because the resulting equation contains only one variable, you can solve for a directly.

$$a=\frac{2435}{7}-2$$

Add -2 to \frac{2435}{7}.

$$a=\frac{2421}{7}$$

The system is now solved.

$$a=\frac{2421}{7},b=\frac{2435}{7}$$

Steps Using Matrices

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

$$a-b=-2,3a+4b=2429$$

Write the equations in matrix form.

$$\left(\begin{matrix}1&-1\\3&4\end{matrix}\right)\left(\begin{matrix}a\\b\end{matrix}\right)=\left(\begin{matrix}-2\\2429\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}1&-1\\3&4\end{matrix}\right).

$$inverse(\left(\begin{matrix}1&-1\\3&4\end{matrix}\right))\left(\begin{matrix}1&-1\\3&4\end{matrix}\right)\left(\begin{matrix}a\\b\end{matrix}\right)=inverse(\left(\begin{matrix}1&-1\\3&4\end{matrix}\right))\left(\begin{matrix}-2\\2429\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}a\\b\end{matrix}\right)=inverse(\left(\begin{matrix}1&-1\\3&4\end{matrix}\right))\left(\begin{matrix}-2\\2429\end{matrix}\right)$$

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

$$\left(\begin{matrix}a\\b\end{matrix}\right)=inverse(\left(\begin{matrix}1&-1\\3&4\end{matrix}\right))\left(\begin{matrix}-2\\2429\end{matrix}\right)$$

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.

$$\left(\begin{matrix}a\\b\end{matrix}\right)=\left(\begin{matrix}\frac{4}{4-\left(-3\right)}&-\frac{-1}{4-\left(-3\right)}\\-\frac{3}{4-\left(-3\right)}&\frac{1}{4-\left(-3\right)}\end{matrix}\right)\left(\begin{matrix}-2\\2429\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}a\\b\end{matrix}\right)=\left(\begin{matrix}\frac{4}{7}&\frac{1}{7}\\-\frac{3}{7}&\frac{1}{7}\end{matrix}\right)\left(\begin{matrix}-2\\2429\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}a\\b\end{matrix}\right)=\left(\begin{matrix}\frac{4}{7}\left(-2\right)+\frac{1}{7}\times 2429\\-\frac{3}{7}\left(-2\right)+\frac{1}{7}\times 2429\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}a\\b\end{matrix}\right)=\left(\begin{matrix}\frac{2421}{7}\\\frac{2435}{7}\end{matrix}\right)$$

Extract the matrix elements a and b.

$$a=\frac{2421}{7},b=\frac{2435}{7}$$

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.

$$a-b=-2,3a+4b=2429$$

To make a and 3a equal, multiply all terms on each side of the first equation by 3 and all terms on each side of the second by 1.

$$3a+3\left(-1\right)b=3\left(-2\right),3a+4b=2429$$

Simplify.

$$3a-3b=-6,3a+4b=2429$$

Subtract 3a+4b=2429 from 3a-3b=-6 by subtracting like terms on each side of the equal sign.

$$3a-3a-3b-4b=-6-2429$$

Add 3a to -3a. Terms 3a and -3a cancel out, leaving an equation with only one variable that can be solved.

$$-3b-4b=-6-2429$$

Add -3b to -4b.

$$-7b=-6-2429$$

Add -6 to -2429.

$$-7b=-2435$$

Divide both sides by -7.

$$b=\frac{2435}{7}$$

Substitute \frac{2435}{7} for b in 3a+4b=2429. Because the resulting equation contains only one variable, you can solve for a directly.

$$3a+4\times \frac{2435}{7}=2429$$

Multiply 4 times \frac{2435}{7}.

$$3a+\frac{9740}{7}=2429$$

Subtract \frac{9740}{7} from both sides of the equation.

$$3a=\frac{7263}{7}$$

Divide both sides by 3.

$$a=\frac{2421}{7}$$

The system is now solved.

$$a=\frac{2421}{7},b=\frac{2435}{7}$$