Solve {l}{a_{1}+9d=4}{15a_{1}+15*14/2d=30}

$$\left\{ \begin{array} { l } { a _ { 1 } + 9 d = 4 } \\ { 15 a _ { 1 } + \frac { 15 \times 14 } { 2 } d = 30 } \end{array} \right.$$

Solve for a_1, d

$$a_{1}=-5$$

$$d=1$$

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.

$$a_{1}+9d=4,15a_{1}+105d=30$$

Choose one of the equations and solve it for a_{1} by isolating a_{1} on the left hand side of the equal sign.

$$a_{1}+9d=4$$

Subtract 9d from both sides of the equation.

$$a_{1}=-9d+4$$

Substitute -9d+4 for a_{1} in the other equation, 15a_{1}+105d=30.

$$15\left(-9d+4\right)+105d=30$$

Multiply 15 times -9d+4.

$$-135d+60+105d=30$$

Add -135d to 105d.

$$-30d+60=30$$

Subtract 60 from both sides of the equation.

$$-30d=-30$$

Divide both sides by -30.

$$d=1$$

Substitute 1 for d in a_{1}=-9d+4. Because the resulting equation contains only one variable, you can solve for a_{1} directly.

$$a_{1}=-9+4$$

Add 4 to -9.

$$a_{1}=-5$$

The system is now solved.

$$a_{1}=-5,d=1$$

Steps Using Matrices

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

$$a_{1}+9d=4,15a_{1}+105d=30$$

Write the equations in matrix form.

$$\left(\begin{matrix}1&9\\15&105\end{matrix}\right)\left(\begin{matrix}a_{1}\\d\end{matrix}\right)=\left(\begin{matrix}4\\30\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}1&9\\15&105\end{matrix}\right).

$$inverse(\left(\begin{matrix}1&9\\15&105\end{matrix}\right))\left(\begin{matrix}1&9\\15&105\end{matrix}\right)\left(\begin{matrix}a_{1}\\d\end{matrix}\right)=inverse(\left(\begin{matrix}1&9\\15&105\end{matrix}\right))\left(\begin{matrix}4\\30\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}a_{1}\\d\end{matrix}\right)=inverse(\left(\begin{matrix}1&9\\15&105\end{matrix}\right))\left(\begin{matrix}4\\30\end{matrix}\right)$$

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

$$\left(\begin{matrix}a_{1}\\d\end{matrix}\right)=inverse(\left(\begin{matrix}1&9\\15&105\end{matrix}\right))\left(\begin{matrix}4\\30\end{matrix}\right)$$

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.

$$\left(\begin{matrix}a_{1}\\d\end{matrix}\right)=\left(\begin{matrix}\frac{105}{105-9\times 15}&-\frac{9}{105-9\times 15}\\-\frac{15}{105-9\times 15}&\frac{1}{105-9\times 15}\end{matrix}\right)\left(\begin{matrix}4\\30\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}a_{1}\\d\end{matrix}\right)=\left(\begin{matrix}-\frac{7}{2}&\frac{3}{10}\\\frac{1}{2}&-\frac{1}{30}\end{matrix}\right)\left(\begin{matrix}4\\30\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}a_{1}\\d\end{matrix}\right)=\left(\begin{matrix}-\frac{7}{2}\times 4+\frac{3}{10}\times 30\\\frac{1}{2}\times 4-\frac{1}{30}\times 30\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}a_{1}\\d\end{matrix}\right)=\left(\begin{matrix}-5\\1\end{matrix}\right)$$

Extract the matrix elements a_{1} and d.

$$a_{1}=-5,d=1$$

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.

$$a_{1}+9d=4,15a_{1}+105d=30$$

To make a_{1} and 15a_{1} equal, multiply all terms on each side of the first equation by 15 and all terms on each side of the second by 1.

$$15a_{1}+15\times 9d=15\times 4,15a_{1}+105d=30$$

Simplify.

$$15a_{1}+135d=60,15a_{1}+105d=30$$

Subtract 15a_{1}+105d=30 from 15a_{1}+135d=60 by subtracting like terms on each side of the equal sign.

$$15a_{1}-15a_{1}+135d-105d=60-30$$

Add 15a_{1} to -15a_{1}. Terms 15a_{1} and -15a_{1} cancel out, leaving an equation with only one variable that can be solved.

$$135d-105d=60-30$$

Add 135d to -105d.

$$30d=60-30$$

Add 60 to -30.

$$30d=30$$

Divide both sides by 30.

$$d=1$$

Substitute 1 for d in 15a_{1}+105d=30. Because the resulting equation contains only one variable, you can solve for a_{1} directly.

$$15a_{1}+105=30$$

Subtract 105 from both sides of the equation.

$$15a_{1}=-75$$

Divide both sides by 15.

$$a_{1}=-5$$

The system is now solved.

$$a_{1}=-5,d=1$$