Question

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

Answer 1

$36y^{2} +24y+4-27=16y^{2} -81+20y^{2} -70y+8y-28\\36y^{2} +24y-23=36y^{2}-109-62y\\24y-23=-109-62y\\24y-23+109+62y=0\\y+1=0\\y=-1$

Answer 2

To solve the equation 4(3y+1)^2 - 27 = (4y+9)(4y-9) + 2(5y+2)(2y-7), we can start by expanding the squares and simplifying the expressions on both sides.

Expanding the left side: 4(3y+1)^2 - 27 = 4(9y^2 + 6y + 1) - 27 = 36y^2 + 24y + 4 - 27 = 36y^2 + 24y - 23

Expanding the right side: (4y+9)(4y-9) + 2(5y+2)(2y-7) = (16y^2 - 81) + 2(10y^2 - 29y - 14) = 16y^2 - 81 + 20y^2 - 58y - 28

Combining like terms on both sides: 36y^2 + 24y - 23 = 16y^2 - 81 + 20y^2 - 58y - 28

Now, let's simplify further by combining like terms: 36y^2 + 24y - 23 = 16y^2 + 20y^2 - 58y - 81 - 28

Combining like terms again: 36y^2 + 24y - 23 = 36y^2 - 58y - 109

To isolate the variable terms on one side, we can subtract 36y^2 from both sides: 24y - 23 = -58y - 109

Next, we'll add 58y to both sides: 82y - 23 = -109

Now, let's isolate the constant term by adding 23 to both sides: 82y = -109 + 23

Simplifying further: 82y = -86

Finally, we can solve for y by dividing both sides by 82: y = -86/82

Simplifying the fraction gives us the solution: y ≈ -1.048