Solve 4(3y+1)^2-27=(4y+9)(4y-9)+2(5y+2)(2y-7)

$$4 ( 3 y + 1 ) ^ { 2 } - 27 = ( 4 y + 9 ) ( 4 y - 9 ) + 2 ( 5 y + 2 ) ( 2 y - 7 )$$

Solve for y

$$y=-1$$

Steps for Solving Linear Equation

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(3y+1\right)^{2}.

$$4\left(9y^{2}+6y+1\right)-27=\left(4y+9\right)\left(4y-9\right)+2\left(5y+2\right)\left(2y-7\right)$$

Use the distributive property to multiply 4 by 9y^{2}+6y+1.

$$36y^{2}+24y+4-27=\left(4y+9\right)\left(4y-9\right)+2\left(5y+2\right)\left(2y-7\right)$$

Subtract 27 from 4 to get -23.

$$36y^{2}+24y-23=\left(4y+9\right)\left(4y-9\right)+2\left(5y+2\right)\left(2y-7\right)$$

Consider \left(4y+9\right)\left(4y-9\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 9.

$$36y^{2}+24y-23=\left(4y\right)^{2}-81+2\left(5y+2\right)\left(2y-7\right)$$

Expand \left(4y\right)^{2}.

$$36y^{2}+24y-23=4^{2}y^{2}-81+2\left(5y+2\right)\left(2y-7\right)$$

Calculate 4 to the power of 2 and get 16.

$$36y^{2}+24y-23=16y^{2}-81+2\left(5y+2\right)\left(2y-7\right)$$

Use the distributive property to multiply 2 by 5y+2.

$$36y^{2}+24y-23=16y^{2}-81+\left(10y+4\right)\left(2y-7\right)$$

Use the distributive property to multiply 10y+4 by 2y-7 and combine like terms.

$$36y^{2}+24y-23=16y^{2}-81+20y^{2}-62y-28$$

Combine 16y^{2} and 20y^{2} to get 36y^{2}.

$$36y^{2}+24y-23=36y^{2}-81-62y-28$$

Subtract 28 from -81 to get -109.

$$36y^{2}+24y-23=36y^{2}-109-62y$$

Subtract 36y^{2} from both sides.

$$36y^{2}+24y-23-36y^{2}=-109-62y$$

Combine 36y^{2} and -36y^{2} to get 0.

$$24y-23=-109-62y$$

Add 62y to both sides.

$$24y-23+62y=-109$$

Combine 24y and 62y to get 86y.

$$86y-23=-109$$

Add 23 to both sides.

$$86y=-109+23$$

Add -109 and 23 to get -86.

$$86y=-86$$

Divide both sides by 86.

$$y=\frac{-86}{86}$$

Divide -86 by 86 to get -1.

$$y=-1$$