Solve {l}{(y+2)^2/3=y+2+y^2/3}{(t+1)(t-1)=(t+2)(t)}

$$\left. \begin{array} { l } { \frac { ( y + 2 ) ^ { 2 } } { 3 } = y + 2 + \frac { y ^ { 2 } } { 3 } } \\ { ( t + 1 ) ( t - 1 ) = ( t + 2 ) ( t ) } \end{array} \right.$$

Solve for y, t

$$y=2$$

$$t=-\frac{1}{2}=-0.5$$

Solution Steps

Consider the first equation. Multiply both sides of the equation by 3.

$$\left(y+2\right)^{2}=3y+6+y^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(y+2\right)^{2}.

$$y^{2}+4y+4=3y+6+y^{2}$$

Subtract 3y from both sides.

$$y^{2}+4y+4-3y=6+y^{2}$$

Combine 4y and -3y to get y.

$$y^{2}+y+4=6+y^{2}$$

Subtract y^{2} from both sides.

$$y^{2}+y+4-y^{2}=6$$

Combine y^{2} and -y^{2} to get 0.

$$y+4=6$$

Subtract 4 from both sides.

$$y=6-4$$

Subtract 4 from 6 to get 2.

$$y=2$$

Consider the second equation. Consider \left(t+1\right)\left(t-1\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 1.

$$t^{2}-1=\left(t+2\right)t$$

Use the distributive property to multiply t+2 by t.

$$t^{2}-1=t^{2}+2t$$

Subtract t^{2} from both sides.

$$t^{2}-1-t^{2}=2t$$

Combine t^{2} and -t^{2} to get 0.

$$-1=2t$$

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

$$2t=-1$$

Divide both sides by 2.

$$t=-\frac{1}{2}$$

The system is now solved.

$$y=2 t=-\frac{1}{2}$$