Solve {l}{text{(1)}34=a*30^2+b*30+50}{text{(2)}0=a*49^2+b*49+50}

$$\left. \begin{array} { l } { \text { (1) } 34 = a \cdot 30 ^ { 2 } + b \cdot 30 + 50 } \\ { \text { (2) } 0 = a \cdot 49 ^ { 2 } + b \cdot 49 + 50 } \end{array} \right.$$

Solve for a, b

$$a=-\frac{358}{13965}\approx -0.025635517$$

$$b=\frac{3292}{13965}\approx 0.235732188$$

Steps Using Substitution

Consider the first equation. Calculate 30 to the power of 2 and get 900.

$$34=a\times 900+b\times 30+50$$

Swap sides so that all variable terms are on the left hand side.

$$a\times 900+b\times 30+50=34$$

Subtract 50 from both sides.

$$a\times 900+b\times 30=34-50$$

Subtract 50 from 34 to get -16.

$$a\times 900+b\times 30=-16$$

Consider the second equation. Calculate 49 to the power of 2 and get 2401.

$$0=a\times 2401+b\times 49+50$$

Swap sides so that all variable terms are on the left hand side.

$$a\times 2401+b\times 49+50=0$$

Subtract 50 from both sides. Anything subtracted from zero gives its negation.

$$a\times 2401+b\times 49=-50$$

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.

$$900a+30b=-16,2401a+49b=-50$$

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

$$900a+30b=-16$$

Subtract 30b from both sides of the equation.

$$900a=-30b-16$$

Divide both sides by 900.

$$a=\frac{1}{900}\left(-30b-16\right)$$

Multiply \frac{1}{900} times -30b-16.

$$a=-\frac{1}{30}b-\frac{4}{225}$$

Substitute -\frac{b}{30}-\frac{4}{225} for a in the other equation, 2401a+49b=-50.

$$2401\left(-\frac{1}{30}b-\frac{4}{225}\right)+49b=-50$$

Multiply 2401 times -\frac{b}{30}-\frac{4}{225}.

$$-\frac{2401}{30}b-\frac{9604}{225}+49b=-50$$

Add -\frac{2401b}{30} to 49b.

$$-\frac{931}{30}b-\frac{9604}{225}=-50$$

Add \frac{9604}{225} to both sides of the equation.

$$-\frac{931}{30}b=-\frac{1646}{225}$$

Divide both sides of the equation by -\frac{931}{30}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$b=\frac{3292}{13965}$$

Substitute \frac{3292}{13965} for b in a=-\frac{1}{30}b-\frac{4}{225}. Because the resulting equation contains only one variable, you can solve for a directly.

$$a=-\frac{1}{30}\times \frac{3292}{13965}-\frac{4}{225}$$

Multiply -\frac{1}{30} times \frac{3292}{13965} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$a=-\frac{1646}{209475}-\frac{4}{225}$$

Add -\frac{4}{225} to -\frac{1646}{209475} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$a=-\frac{358}{13965}$$

The system is now solved.

$$a=-\frac{358}{13965},b=\frac{3292}{13965}$$

Steps Using Matrices

Consider the first equation. Calculate 30 to the power of 2 and get 900.

$$34=a\times 900+b\times 30+50$$

Swap sides so that all variable terms are on the left hand side.

$$a\times 900+b\times 30+50=34$$

Subtract 50 from both sides.

$$a\times 900+b\times 30=34-50$$

Subtract 50 from 34 to get -16.

$$a\times 900+b\times 30=-16$$

Consider the second equation. Calculate 49 to the power of 2 and get 2401.

$$0=a\times 2401+b\times 49+50$$

Swap sides so that all variable terms are on the left hand side.

$$a\times 2401+b\times 49+50=0$$

Subtract 50 from both sides. Anything subtracted from zero gives its negation.

$$a\times 2401+b\times 49=-50$$

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

$$900a+30b=-16,2401a+49b=-50$$

Write the equations in matrix form.

$$\left(\begin{matrix}900&30\\2401&49\end{matrix}\right)\left(\begin{matrix}a\\b\end{matrix}\right)=\left(\begin{matrix}-16\\-50\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}900&30\\2401&49\end{matrix}\right).

$$inverse(\left(\begin{matrix}900&30\\2401&49\end{matrix}\right))\left(\begin{matrix}900&30\\2401&49\end{matrix}\right)\left(\begin{matrix}a\\b\end{matrix}\right)=inverse(\left(\begin{matrix}900&30\\2401&49\end{matrix}\right))\left(\begin{matrix}-16\\-50\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}a\\b\end{matrix}\right)=inverse(\left(\begin{matrix}900&30\\2401&49\end{matrix}\right))\left(\begin{matrix}-16\\-50\end{matrix}\right)$$

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

$$\left(\begin{matrix}a\\b\end{matrix}\right)=inverse(\left(\begin{matrix}900&30\\2401&49\end{matrix}\right))\left(\begin{matrix}-16\\-50\end{matrix}\right)$$

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.

$$\left(\begin{matrix}a\\b\end{matrix}\right)=\left(\begin{matrix}\frac{49}{900\times 49-30\times 2401}&-\frac{30}{900\times 49-30\times 2401}\\-\frac{2401}{900\times 49-30\times 2401}&\frac{900}{900\times 49-30\times 2401}\end{matrix}\right)\left(\begin{matrix}-16\\-50\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}a\\b\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{570}&\frac{1}{931}\\\frac{49}{570}&-\frac{30}{931}\end{matrix}\right)\left(\begin{matrix}-16\\-50\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}a\\b\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{570}\left(-16\right)+\frac{1}{931}\left(-50\right)\\\frac{49}{570}\left(-16\right)-\frac{30}{931}\left(-50\right)\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}a\\b\end{matrix}\right)=\left(\begin{matrix}-\frac{358}{13965}\\\frac{3292}{13965}\end{matrix}\right)$$

Extract the matrix elements a and b.

$$a=-\frac{358}{13965},b=\frac{3292}{13965}$$

Steps Using Elimination

Consider the first equation. Calculate 30 to the power of 2 and get 900.

$$34=a\times 900+b\times 30+50$$

Swap sides so that all variable terms are on the left hand side.

$$a\times 900+b\times 30+50=34$$

Subtract 50 from both sides.

$$a\times 900+b\times 30=34-50$$

Subtract 50 from 34 to get -16.

$$a\times 900+b\times 30=-16$$

Consider the second equation. Calculate 49 to the power of 2 and get 2401.

$$0=a\times 2401+b\times 49+50$$

Swap sides so that all variable terms are on the left hand side.

$$a\times 2401+b\times 49+50=0$$

Subtract 50 from both sides. Anything subtracted from zero gives its negation.

$$a\times 2401+b\times 49=-50$$

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.

$$900a+30b=-16,2401a+49b=-50$$

To make 900a and 2401a equal, multiply all terms on each side of the first equation by 2401 and all terms on each side of the second by 900.

$$2401\times 900a+2401\times 30b=2401\left(-16\right),900\times 2401a+900\times 49b=900\left(-50\right)$$

Simplify.

$$2160900a+72030b=-38416,2160900a+44100b=-45000$$

Subtract 2160900a+44100b=-45000 from 2160900a+72030b=-38416 by subtracting like terms on each side of the equal sign.

$$2160900a-2160900a+72030b-44100b=-38416+45000$$

Add 2160900a to -2160900a. Terms 2160900a and -2160900a cancel out, leaving an equation with only one variable that can be solved.

$$72030b-44100b=-38416+45000$$

Add 72030b to -44100b.

$$27930b=-38416+45000$$

Add -38416 to 45000.

$$27930b=6584$$

Divide both sides by 27930.

$$b=\frac{3292}{13965}$$

Substitute \frac{3292}{13965} for b in 2401a+49b=-50. Because the resulting equation contains only one variable, you can solve for a directly.

$$2401a+49\times \frac{3292}{13965}=-50$$

Multiply 49 times \frac{3292}{13965}.

$$2401a+\frac{3292}{285}=-50$$

Subtract \frac{3292}{285} from both sides of the equation.

$$2401a=-\frac{17542}{285}$$

Divide both sides by 2401.

$$a=-\frac{358}{13965}$$

The system is now solved.

$$a=-\frac{358}{13965},b=\frac{3292}{13965}$$