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question 8 - test #2 bonus analyze how would you solve 5x = 35? how wou…

Question

question 8 - test #2 bonus
analyze how would you solve 5x = 35? how would you solve 5 + x = 35? how are the methods similar and how are they different?
to solve 5x = 35, i would select choice to get x = select choice. to solve 5 + x = 35, i would select choice to get x = select choice. in both equations, i used properties of equality to isolate the variable. i used the select choice property of equality in the first equation and i used the select choice property of equality in the second equation.

Explanation:

Step1: Solve 5x = 35

Divide both sides by 5. So, $\frac{5x}{5}=\frac{35}{5}$, which gives $x = 7$.

Step2: Solve 5 + x = 35

Subtract 5 from both sides. So, $5 + x-5=35 - 5$, which gives $x = 30$.

Step3: Analyze similarities

In both, we used properties of equality to isolate x.

Step4: Analyze differences

For $5x = 35$, we used the Division Property of Equality. For $5 + x = 35$, we used the Subtraction Property of Equality.

Answer:

To solve $5x = 35$, I would divide both sides by 5 to get $x = 7$. To solve $5 + x = 35$, I would subtract 5 from both sides to get $x = 30$. In both equations, I used properties of equality to isolate the variable. I used the Division Property of Equality in the first equation and I used the Subtraction Property of Equality in the second equation.