How do you solve $4 (x - 2) = x + 7$ $4( x - 2) = x + 7$ ?

Question

How do you solve $4 (x - 2) = x + 7$ $4( x - 2) = x + 7$ ?

2 Answers

Impact of this question

1950 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-06-02T02:56:24

$x = 5$ $x=5$

Explanation:

Given: $4 (x - 2) = x + 7$ $4(x-2)=x+7$ .

Expand the left side.

$4 x - 8 = x + 7$ $4x-8=x+7$

Subtract $7$ $7$ from both sides.

$4 x - 8 - 7 = x$ $4x-8-7=x$

$4 x - 15 = x$ $4x-15=x$

Subtract $4 x$ $4x$ from both sides.

$- 15 = x - 4 x$ $-15=x-4x$

$- 15 = - 3 x$ $-15=-3x$

Divide by $- 3$ $-3$ .

$x = \frac{- 15}{- 3}$ $x=(-15)/-3$

$= 5$ $=5$

Answer 2 · 2018-06-02T03:03:28

$x = 5$ $x=5$

Explanation:

To solve for the variable in the equation $4 (x - 2) = x + 7$ $4(x-2)=x+7$

Begin by eliminating the parenthesis by using the distributive property.

$4 (x - 2) = x + 7$ $4(x-2)=x+7$

$4 x - 8 = x + 7$ $4x-8=x+7$

$3 x - 8 = 7$ $3x-8=7$

Use the additive inverse to get the $x$ $x$ terms on one side of the equation.

$4x-8-x= cancelx +7 cancel(-x)$

$3x-8=7$

Use the additive inverse to get the constant values on the same side of the equation

$3xcancel(-8) cancel(+8)=7+8$

$3x=15$

Now use the multiplicative inverse to solve for $x$

$(cancel(3)x)/cancel3 =15/3$

$x=5$

Science

Math

Humanities

... and beyond

How do you solve $4 (x - 2) = x + 7$ $4( x - 2) = x + 7$ ?

2 Answers

Explanation:

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you solve 4( x - 2) = x + 74(x−2)=x+74( x - 2) = x + 7?

2 Answers

Explanation:

Explanation:

Related questions

Impact of this question

How do you solve $4 (x - 2) = x + 7$ $4( x - 2) = x + 7$ ?