Solve {l}{2a+b+10=170}{a+2b+30=170}

$$\left\{ \begin{array} { l } { 2 a + b + 10 = 170 } \\ { a + 2 b + 30 = 170 } \end{array} \right.$$

Solve for a, b

$$a=60$$

$$b=40$$

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.

$$2a+b+10=170,a+2b+30=170$$

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

$$2a+b+10=170$$

Subtract 10 from both sides of the equation.

$$2a+b=160$$

Subtract b from both sides of the equation.

$$2a=-b+160$$

Divide both sides by 2.

$$a=\frac{1}{2}\left(-b+160\right)$$

Multiply \frac{1}{2} times -b+160.

$$a=-\frac{1}{2}b+80$$

Substitute -\frac{b}{2}+80 for a in the other equation, a+2b+30=170.

$$-\frac{1}{2}b+80+2b+30=170$$

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

$$\frac{3}{2}b+80+30=170$$

Add 80 to 30.

$$\frac{3}{2}b+110=170$$

Subtract 110 from both sides of the equation.

$$\frac{3}{2}b=60$$

Divide both sides of the equation by \frac{3}{2}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$b=40$$

Substitute 40 for b in a=-\frac{1}{2}b+80. Because the resulting equation contains only one variable, you can solve for a directly.

$$a=-\frac{1}{2}\times 40+80$$

Multiply -\frac{1}{2} times 40.

$$a=-20+80$$

Add 80 to -20.

$$a=60$$

The system is now solved.

$$a=60,b=40$$

Steps Using Matrices

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

$$2a+b+10=170,a+2b+30=170$$

Write the equations in matrix form.

$$\left(\begin{matrix}2&1\\1&2\end{matrix}\right)\left(\begin{matrix}a\\b\end{matrix}\right)=\left(\begin{matrix}160\\140\end{matrix}\right)$$

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

$$inverse(\left(\begin{matrix}2&1\\1&2\end{matrix}\right))\left(\begin{matrix}2&1\\1&2\end{matrix}\right)\left(\begin{matrix}a\\b\end{matrix}\right)=inverse(\left(\begin{matrix}2&1\\1&2\end{matrix}\right))\left(\begin{matrix}160\\140\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}a\\b\end{matrix}\right)=inverse(\left(\begin{matrix}2&1\\1&2\end{matrix}\right))\left(\begin{matrix}160\\140\end{matrix}\right)$$

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

$$\left(\begin{matrix}a\\b\end{matrix}\right)=inverse(\left(\begin{matrix}2&1\\1&2\end{matrix}\right))\left(\begin{matrix}160\\140\end{matrix}\right)$$

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.

$$\left(\begin{matrix}a\\b\end{matrix}\right)=\left(\begin{matrix}\frac{2}{2\times 2-1}&-\frac{1}{2\times 2-1}\\-\frac{1}{2\times 2-1}&\frac{2}{2\times 2-1}\end{matrix}\right)\left(\begin{matrix}160\\140\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}a\\b\end{matrix}\right)=\left(\begin{matrix}\frac{2}{3}&-\frac{1}{3}\\-\frac{1}{3}&\frac{2}{3}\end{matrix}\right)\left(\begin{matrix}160\\140\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}a\\b\end{matrix}\right)=\left(\begin{matrix}\frac{2}{3}\times 160-\frac{1}{3}\times 140\\-\frac{1}{3}\times 160+\frac{2}{3}\times 140\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}a\\b\end{matrix}\right)=\left(\begin{matrix}60\\40\end{matrix}\right)$$

Extract the matrix elements a and b.

$$a=60,b=40$$

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.

$$2a+b+10=170,a+2b+30=170$$

To make 2a and a equal, multiply all terms on each side of the first equation by 1 and all terms on each side of the second by 2.

$$2a+b+10=170,2a+2\times 2b+2\times 30=2\times 170$$

Simplify.

$$2a+b+10=170,2a+4b+60=340$$

Subtract 2a+4b+60=340 from 2a+b+10=170 by subtracting like terms on each side of the equal sign.

$$2a-2a+b-4b+10-60=170-340$$

Add 2a to -2a. Terms 2a and -2a cancel out, leaving an equation with only one variable that can be solved.

$$b-4b+10-60=170-340$$

Add b to -4b.

$$-3b+10-60=170-340$$

Add 10 to -60.

$$-3b-50=170-340$$

Add 170 to -340.

$$-3b-50=-170$$

Add 50 to both sides of the equation.

$$-3b=-120$$

Divide both sides by -3.

$$b=40$$

Substitute 40 for b in a+2b+30=170. Because the resulting equation contains only one variable, you can solve for a directly.

$$a+2\times 40+30=170$$

Multiply 2 times 40.

$$a+80+30=170$$

Add 80 to 30.

$$a+110=170$$

Subtract 110 from both sides of the equation.

$$a=60$$

The system is now solved.

$$a=60,b=40$$