Solve {l}{4a_{1}+6d=3}{3a_{1}+21d=4}

$$\left. \begin{array} { l } { 4 a _ { 1 } + 6 d = 3 } \\ { 3 a _ { 1 } + 21 d = 4 } \end{array} \right.$$

Solve for a_1, d

$$a_{1}=\frac{13}{22}\approx 0.590909091$$

$$d=\frac{7}{66}\approx 0.106060606$$

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.

$$4a_{1}+6d=3,3a_{1}+21d=4$$

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

$$4a_{1}+6d=3$$

Subtract 6d from both sides of the equation.

$$4a_{1}=-6d+3$$

Divide both sides by 4.

$$a_{1}=\frac{1}{4}\left(-6d+3\right)$$

Multiply \frac{1}{4} times -6d+3.

$$a_{1}=-\frac{3}{2}d+\frac{3}{4}$$

Substitute -\frac{3d}{2}+\frac{3}{4} for a_{1} in the other equation, 3a_{1}+21d=4.

$$3\left(-\frac{3}{2}d+\frac{3}{4}\right)+21d=4$$

Multiply 3 times -\frac{3d}{2}+\frac{3}{4}.

$$-\frac{9}{2}d+\frac{9}{4}+21d=4$$

Add -\frac{9d}{2} to 21d.

$$\frac{33}{2}d+\frac{9}{4}=4$$

Subtract \frac{9}{4} from both sides of the equation.

$$\frac{33}{2}d=\frac{7}{4}$$

Divide both sides of the equation by \frac{33}{2}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$d=\frac{7}{66}$$

Substitute \frac{7}{66} for d in a_{1}=-\frac{3}{2}d+\frac{3}{4}. Because the resulting equation contains only one variable, you can solve for a_{1} directly.

$$a_{1}=-\frac{3}{2}\times \frac{7}{66}+\frac{3}{4}$$

Multiply -\frac{3}{2} times \frac{7}{66} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$a_{1}=-\frac{7}{44}+\frac{3}{4}$$

Add \frac{3}{4} to -\frac{7}{44} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$a_{1}=\frac{13}{22}$$

The system is now solved.

$$a_{1}=\frac{13}{22},d=\frac{7}{66}$$

Steps Using Matrices

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

$$4a_{1}+6d=3,3a_{1}+21d=4$$

Write the equations in matrix form.

$$\left(\begin{matrix}4&6\\3&21\end{matrix}\right)\left(\begin{matrix}a_{1}\\d\end{matrix}\right)=\left(\begin{matrix}3\\4\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}4&6\\3&21\end{matrix}\right).

$$inverse(\left(\begin{matrix}4&6\\3&21\end{matrix}\right))\left(\begin{matrix}4&6\\3&21\end{matrix}\right)\left(\begin{matrix}a_{1}\\d\end{matrix}\right)=inverse(\left(\begin{matrix}4&6\\3&21\end{matrix}\right))\left(\begin{matrix}3\\4\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}a_{1}\\d\end{matrix}\right)=inverse(\left(\begin{matrix}4&6\\3&21\end{matrix}\right))\left(\begin{matrix}3\\4\end{matrix}\right)$$

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

$$\left(\begin{matrix}a_{1}\\d\end{matrix}\right)=inverse(\left(\begin{matrix}4&6\\3&21\end{matrix}\right))\left(\begin{matrix}3\\4\end{matrix}\right)$$

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.

$$\left(\begin{matrix}a_{1}\\d\end{matrix}\right)=\left(\begin{matrix}\frac{21}{4\times 21-6\times 3}&-\frac{6}{4\times 21-6\times 3}\\-\frac{3}{4\times 21-6\times 3}&\frac{4}{4\times 21-6\times 3}\end{matrix}\right)\left(\begin{matrix}3\\4\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}a_{1}\\d\end{matrix}\right)=\left(\begin{matrix}\frac{7}{22}&-\frac{1}{11}\\-\frac{1}{22}&\frac{2}{33}\end{matrix}\right)\left(\begin{matrix}3\\4\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}a_{1}\\d\end{matrix}\right)=\left(\begin{matrix}\frac{7}{22}\times 3-\frac{1}{11}\times 4\\-\frac{1}{22}\times 3+\frac{2}{33}\times 4\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}a_{1}\\d\end{matrix}\right)=\left(\begin{matrix}\frac{13}{22}\\\frac{7}{66}\end{matrix}\right)$$

Extract the matrix elements a_{1} and d.

$$a_{1}=\frac{13}{22},d=\frac{7}{66}$$

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.

$$4a_{1}+6d=3,3a_{1}+21d=4$$

To make 4a_{1} and 3a_{1} equal, multiply all terms on each side of the first equation by 3 and all terms on each side of the second by 4.

$$3\times 4a_{1}+3\times 6d=3\times 3,4\times 3a_{1}+4\times 21d=4\times 4$$

Simplify.

$$12a_{1}+18d=9,12a_{1}+84d=16$$

Subtract 12a_{1}+84d=16 from 12a_{1}+18d=9 by subtracting like terms on each side of the equal sign.

$$12a_{1}-12a_{1}+18d-84d=9-16$$

Add 12a_{1} to -12a_{1}. Terms 12a_{1} and -12a_{1} cancel out, leaving an equation with only one variable that can be solved.

$$18d-84d=9-16$$

Add 18d to -84d.

$$-66d=9-16$$

Add 9 to -16.

$$-66d=-7$$

Divide both sides by -66.

$$d=\frac{7}{66}$$

Substitute \frac{7}{66} for d in 3a_{1}+21d=4. Because the resulting equation contains only one variable, you can solve for a_{1} directly.

$$3a_{1}+21\times \frac{7}{66}=4$$

Multiply 21 times \frac{7}{66}.

$$3a_{1}+\frac{49}{22}=4$$

Subtract \frac{49}{22} from both sides of the equation.

$$3a_{1}=\frac{39}{22}$$

Divide both sides by 3.

$$a_{1}=\frac{13}{22}$$

The system is now solved.

$$a_{1}=\frac{13}{22},d=\frac{7}{66}$$