Solve {l}{7a+21d=49}{17a+136d=289}

$$\left. \begin{array} { l } { 7 a + 21 d = 49 } \\ { 17 a + 136 d = 289 } \end{array} \right.$$

Solve for a, d

$$a=1$$

$$d=2$$

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.

$$7a+21d=49,17a+136d=289$$

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

$$7a+21d=49$$

Subtract 21d from both sides of the equation.

$$7a=-21d+49$$

Divide both sides by 7.

$$a=\frac{1}{7}\left(-21d+49\right)$$

Multiply \frac{1}{7} times -21d+49.

$$a=-3d+7$$

Substitute -3d+7 for a in the other equation, 17a+136d=289.

$$17\left(-3d+7\right)+136d=289$$

Multiply 17 times -3d+7.

$$-51d+119+136d=289$$

Add -51d to 136d.

$$85d+119=289$$

Subtract 119 from both sides of the equation.

$$85d=170$$

Divide both sides by 85.

$$d=2$$

Substitute 2 for d in a=-3d+7. Because the resulting equation contains only one variable, you can solve for a directly.

$$a=-3\times 2+7$$

Multiply -3 times 2.

$$a=-6+7$$

Add 7 to -6.

$$a=1$$

The system is now solved.

$$a=1,d=2$$

Steps Using Matrices

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

$$7a+21d=49,17a+136d=289$$

Write the equations in matrix form.

$$\left(\begin{matrix}7&21\\17&136\end{matrix}\right)\left(\begin{matrix}a\\d\end{matrix}\right)=\left(\begin{matrix}49\\289\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}7&21\\17&136\end{matrix}\right).

$$inverse(\left(\begin{matrix}7&21\\17&136\end{matrix}\right))\left(\begin{matrix}7&21\\17&136\end{matrix}\right)\left(\begin{matrix}a\\d\end{matrix}\right)=inverse(\left(\begin{matrix}7&21\\17&136\end{matrix}\right))\left(\begin{matrix}49\\289\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}a\\d\end{matrix}\right)=inverse(\left(\begin{matrix}7&21\\17&136\end{matrix}\right))\left(\begin{matrix}49\\289\end{matrix}\right)$$

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

$$\left(\begin{matrix}a\\d\end{matrix}\right)=inverse(\left(\begin{matrix}7&21\\17&136\end{matrix}\right))\left(\begin{matrix}49\\289\end{matrix}\right)$$

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.

$$\left(\begin{matrix}a\\d\end{matrix}\right)=\left(\begin{matrix}\frac{136}{7\times 136-21\times 17}&-\frac{21}{7\times 136-21\times 17}\\-\frac{17}{7\times 136-21\times 17}&\frac{7}{7\times 136-21\times 17}\end{matrix}\right)\left(\begin{matrix}49\\289\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}a\\d\end{matrix}\right)=\left(\begin{matrix}\frac{8}{35}&-\frac{3}{85}\\-\frac{1}{35}&\frac{1}{85}\end{matrix}\right)\left(\begin{matrix}49\\289\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}a\\d\end{matrix}\right)=\left(\begin{matrix}\frac{8}{35}\times 49-\frac{3}{85}\times 289\\-\frac{1}{35}\times 49+\frac{1}{85}\times 289\end{matrix}\right)$$

Do the arithmetic.

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

Extract the matrix elements a and d.

$$a=1,d=2$$

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.

$$7a+21d=49,17a+136d=289$$

To make 7a and 17a equal, multiply all terms on each side of the first equation by 17 and all terms on each side of the second by 7.

$$17\times 7a+17\times 21d=17\times 49,7\times 17a+7\times 136d=7\times 289$$

Simplify.

$$119a+357d=833,119a+952d=2023$$

Subtract 119a+952d=2023 from 119a+357d=833 by subtracting like terms on each side of the equal sign.

$$119a-119a+357d-952d=833-2023$$

Add 119a to -119a. Terms 119a and -119a cancel out, leaving an equation with only one variable that can be solved.

$$357d-952d=833-2023$$

Add 357d to -952d.

$$-595d=833-2023$$

Add 833 to -2023.

$$-595d=-1190$$

Divide both sides by -595.

$$d=2$$

Substitute 2 for d in 17a+136d=289. Because the resulting equation contains only one variable, you can solve for a directly.

$$17a+136\times 2=289$$

Multiply 136 times 2.

$$17a+272=289$$

Subtract 272 from both sides of the equation.

$$17a=17$$

Divide both sides by 17.

$$a=1$$

The system is now solved.

$$a=1,d=2$$