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Justifying Steps

Match each step to its justification to solve for [tex]\( x \)[/tex]:

[tex]\[
\begin{array}{rl}
1. & 2x + 5 = 19 \\
2. & 2x = 14 \\
3. & x = 7 \\
\end{array}
\][/tex]

A. Given
B. Subtract 5 from both sides
C. Divide both sides by 2


Sagot :

It appears that there is some confusion in the layout of the equations and operations. Let's break down and correct them step-by-step, ensuring that each step includes a clear justification along the way.

Assume we want to solve for [tex]\(x\)[/tex] in the initial equation [tex]\(2x + 5 = 19\)[/tex].

1. Given:
[tex]\[ 2x + 5 = 19 \][/tex]

2. Subtraction property of equality:
To isolate the term [tex]\(2x\)[/tex], we subtract 5 from both sides of the equation.
[tex]\[ 2x + 5 - 5 = 19 - 5 \][/tex]
Simplifying, we get:
[tex]\[ 2x = 14 \][/tex]

3. Division property of equality:
To solve for [tex]\(x\)[/tex], divide both sides by 2.
[tex]\[ \frac{2x}{2} = \frac{14}{2} \][/tex]
Simplifying, we get:
[tex]\[ x = 7 \][/tex]

Therefore, the solution to the equation [tex]\(2x + 5 = 19\)[/tex] is [tex]\(x = 7\)[/tex]. Matching each step to its justification:

- [tex]\(\boxed{2x + 5 = 19}\)[/tex] - Given
- [tex]\(\boxed{2x + 5 - 5 = 19 - 5}\)[/tex] - Subtraction property of equality
- [tex]\(\boxed{2x = 14}\)[/tex] - Simplified after applying the subtraction property
- [tex]\(\boxed{\frac{2x}{2} = \frac{14}{2}}\)[/tex] - Division property of equality
- [tex]\(\boxed{x = 7}\)[/tex] - Simplified after applying the division property

Thus, each step can be clearly understood and justified within the problem-solving process for linear equations.