Which of the following equations is equivalent to 4[x + 2(3x – 7)] = 22x – 65?28x – 7 = 22x – 6528x – 56 = 22x – 6510x – 14 = 22x – 6516x – 28 = 22x – 65

The equation 4[x + 2(3x – 7)] = 22x – 65 is equivalent to 16x – 28 = 22x – 65. This can be proven by expanding the original equation and demonstrating that both equations are equal.

When we expand 4[x + 2(3x – 7)] we get:
4[x + 6x – 14]
From here, we can use distributive property and simplify this to:
4(7ox – 14)
Then using the commutative property of multiplication, this becomes:
28X – 56 – (Equation 1)

Which of the following equations is equivalent to 4[x + 2(3x – 7)] = 22x – 65?28x – 7 = 22x – 6528x – 56 = 22x – 6510x – 14 = 22x – 6516x – 28 = 22x – 65

Next, let’s look at the second equation given which is: 22X – 65 – (Equation 2)
By subtracting 28 from each side of Equation 1 and then subtracting 22 from each side in Equation 2, you end up with the same answer: 16X – 28 = 22X–65. This proves that both equations are equivalent. By solving for x on either side of the new equation you will find that x = 3/2 or 1.5. Therefore, 16 X – 28 = 22X – 65 is an equivalent equation to our original one which was 4 [ x + 2(3 x – 7)].

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