Solve c=(49/5)^2b+7^2*10adiv34*3

$$c = ( \frac { 49 } { 5 } ) ^ { 2 } b + 7 ^ { 2 } \times 10 a \div 34 \times 3$$

Solve for a

$$a=\frac{17c}{735}-\frac{833b}{375}$$

Steps for Solving Linear Equation

Calculate \frac{49}{5} to the power of 2 and get \frac{2401}{25}.

$$c=\frac{2401}{25}b+\frac{7^{2}\times 10a}{34}\times 3$$

Calculate 7 to the power of 2 and get 49.

$$c=\frac{2401}{25}b+\frac{49\times 10a}{34}\times 3$$

Multiply 49 and 10 to get 490.

$$c=\frac{2401}{25}b+\frac{490a}{34}\times 3$$

Divide 490a by 34 to get \frac{245}{17}a.

$$c=\frac{2401}{25}b+\frac{245}{17}a\times 3$$

Multiply \frac{245}{17} and 3 to get \frac{735}{17}.

$$c=\frac{2401}{25}b+\frac{735}{17}a$$

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

$$\frac{2401}{25}b+\frac{735}{17}a=c$$

Subtract \frac{2401}{25}b from both sides.

$$\frac{735}{17}a=c-\frac{2401}{25}b$$

The equation is in standard form.

$$\frac{735}{17}a=-\frac{2401b}{25}+c$$

Divide both sides of the equation by \frac{735}{17}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$\frac{\frac{735}{17}a}{\frac{735}{17}}=\frac{-\frac{2401b}{25}+c}{\frac{735}{17}}$$

Dividing by \frac{735}{17} undoes the multiplication by \frac{735}{17}.

$$a=\frac{-\frac{2401b}{25}+c}{\frac{735}{17}}$$

Divide c-\frac{2401b}{25} by \frac{735}{17} by multiplying c-\frac{2401b}{25} by the reciprocal of \frac{735}{17}.

$$a=\frac{17c}{735}-\frac{833b}{375}$$

Solve for b

$$b=\frac{25c}{2401}-\frac{375a}{833}$$

Steps for Solving Linear Equation

Calculate \frac{49}{5} to the power of 2 and get \frac{2401}{25}.

$$c=\frac{2401}{25}b+\frac{7^{2}\times 10a}{34}\times 3$$

Calculate 7 to the power of 2 and get 49.

$$c=\frac{2401}{25}b+\frac{49\times 10a}{34}\times 3$$

Multiply 49 and 10 to get 490.

$$c=\frac{2401}{25}b+\frac{490a}{34}\times 3$$

Divide 490a by 34 to get \frac{245}{17}a.

$$c=\frac{2401}{25}b+\frac{245}{17}a\times 3$$

Multiply \frac{245}{17} and 3 to get \frac{735}{17}.

$$c=\frac{2401}{25}b+\frac{735}{17}a$$

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

$$\frac{2401}{25}b+\frac{735}{17}a=c$$

Subtract \frac{735}{17}a from both sides.

$$\frac{2401}{25}b=c-\frac{735}{17}a$$

The equation is in standard form.

$$\frac{2401}{25}b=-\frac{735a}{17}+c$$

Divide both sides of the equation by \frac{2401}{25}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$\frac{\frac{2401}{25}b}{\frac{2401}{25}}=\frac{-\frac{735a}{17}+c}{\frac{2401}{25}}$$

Dividing by \frac{2401}{25} undoes the multiplication by \frac{2401}{25}.

$$b=\frac{-\frac{735a}{17}+c}{\frac{2401}{25}}$$

Divide c-\frac{735a}{17} by \frac{2401}{25} by multiplying c-\frac{735a}{17} by the reciprocal of \frac{2401}{25}.

$$b=\frac{25c}{2401}-\frac{375a}{833}$$