How would you solve the problem 7-4[3-(4y-5)] ?

First expand the brackets; 4*3, 4*-4y, 4*5
So you have 7-(12-16y+20)
Then expand again
So you have 7-12+16y-20
Which simplifies to 16y-25

7 - 4[3 - (4y - 5)]
= 7 - 4[3 - 4y + 5]
= 7 - 4[-4y + 3 + 5]
= 7 - 4[-4y + 8]
= 7 + 4*4y - 4*8
= 7 + 16y - 32
= 16y - 32 + 7
= 16y - 25

7-4[3-(4y-5)]

3 [ 3 - 4y + 5] *multiply - by all number inside ( )

9 -12y +15

-12y +24 = 0 you want to find y, so all equation =0

-12y = -24

y = 2

I hope it's help honey..... good luck...

3[3-4y+5]
9-12y+15
16-2y

7 - 4 (8 - 4y)
7 - 32 + 16y
16y - 25

7 - 4{3-(4y-5)} =
7 - 4(3-4y + 5) =
7 - 12 + 16y - 20 =
16y - 25 =
16y = 20
y = 5/4.

7-4[3-4y+5]
7-12+16y-20
16y-25
16y=25
y=16/25

7-4[3-{4y-5}]
7-4[3-4y+5]
7-12+16y+20
16y+15 hence the answer