Solve frac{a_{1}(1-q^3)}{1-q}=a_{1}*q+10a

$$\frac { a _ { 1 } ( 1 - q ^ { 3 } ) } { 1 - q } = a _ { 1 } \cdot q + 10 a$$

Solve for a

$$a=\frac{a_{1}\left(q^{2}+1\right)}{10}$$

$$q\neq 1$$

Steps for Solving Linear Equation

Multiply both sides of the equation by -q+1.

$$a_{1}\left(1-q^{3}\right)=a_{1}q\left(-q+1\right)+10a\left(-q+1\right)$$

Use the distributive property to multiply a_{1} by 1-q^{3}.

$$a_{1}-a_{1}q^{3}=a_{1}q\left(-q+1\right)+10a\left(-q+1\right)$$

Use the distributive property to multiply a_{1}q by -q+1.

$$a_{1}-a_{1}q^{3}=-a_{1}q^{2}+a_{1}q+10a\left(-q+1\right)$$

Use the distributive property to multiply 10a by -q+1.

$$a_{1}-a_{1}q^{3}=-a_{1}q^{2}+a_{1}q-10aq+10a$$

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

$$-a_{1}q^{2}+a_{1}q-10aq+10a=a_{1}-a_{1}q^{3}$$

Add a_{1}q^{2} to both sides.

$$a_{1}q-10aq+10a=a_{1}-a_{1}q^{3}+a_{1}q^{2}$$

Subtract a_{1}q from both sides.

$$-10aq+10a=a_{1}-a_{1}q^{3}+a_{1}q^{2}-a_{1}q$$

Combine all terms containing a.

$$\left(-10q+10\right)a=a_{1}-a_{1}q^{3}+a_{1}q^{2}-a_{1}q$$

The equation is in standard form.

$$\left(10-10q\right)a=a_{1}-a_{1}q+a_{1}q^{2}-a_{1}q^{3}$$

Divide both sides by -10q+10.

$$\frac{\left(10-10q\right)a}{10-10q}=\frac{a_{1}\left(1-q\right)\left(q^{2}+1\right)}{10-10q}$$

Dividing by -10q+10 undoes the multiplication by -10q+10.

$$a=\frac{a_{1}\left(1-q\right)\left(q^{2}+1\right)}{10-10q}$$

Divide a_{1}\left(1-q\right)\left(1+q^{2}\right) by -10q+10.

$$a=\frac{a_{1}\left(q^{2}+1\right)}{10}$$

Solve for a_1

$$a_{1}=\frac{10a}{q^{2}+1}$$

$$q\neq 1$$

Steps for Solving Linear Equation

Multiply both sides of the equation by -q+1.

$$a_{1}\left(1-q^{3}\right)=a_{1}q\left(-q+1\right)+10a\left(-q+1\right)$$

Use the distributive property to multiply a_{1} by 1-q^{3}.

$$a_{1}-a_{1}q^{3}=a_{1}q\left(-q+1\right)+10a\left(-q+1\right)$$

Use the distributive property to multiply a_{1}q by -q+1.

$$a_{1}-a_{1}q^{3}=-a_{1}q^{2}+a_{1}q+10a\left(-q+1\right)$$

Use the distributive property to multiply 10a by -q+1.

$$a_{1}-a_{1}q^{3}=-a_{1}q^{2}+a_{1}q-10aq+10a$$

Add a_{1}q^{2} to both sides.

$$a_{1}-a_{1}q^{3}+a_{1}q^{2}=a_{1}q-10aq+10a$$

Subtract a_{1}q from both sides.

$$a_{1}-a_{1}q^{3}+a_{1}q^{2}-a_{1}q=-10aq+10a$$

Combine all terms containing a_{1}.

$$\left(1-q^{3}+q^{2}-q\right)a_{1}=-10aq+10a$$

The equation is in standard form.

$$\left(1-q+q^{2}-q^{3}\right)a_{1}=10a-10aq$$

Divide both sides by 1-q^{3}+q^{2}-q.

$$\frac{\left(1-q+q^{2}-q^{3}\right)a_{1}}{1-q+q^{2}-q^{3}}=\frac{10a\left(1-q\right)}{1-q+q^{2}-q^{3}}$$

Dividing by 1-q^{3}+q^{2}-q undoes the multiplication by 1-q^{3}+q^{2}-q.

$$a_{1}=\frac{10a\left(1-q\right)}{1-q+q^{2}-q^{3}}$$

Divide 10a\left(-q+1\right) by 1-q^{3}+q^{2}-q.

$$a_{1}=\frac{10a}{q^{2}+1}$$