Let's tackle each of these problems one by one:

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1. Match each equation with an equivalent equation.

Given equation:
e. $7x = 1$

To solve for $x$, divide both sides by 7:
$x = \frac{1}{7}$ Look for an equation in the answer choices that also simplifies to $x = \frac{1}{7}$.

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2. Mai says that equations A and B have the same solution.

• Equation A: $-3(x + 7) = 24$
• Equation B: $x + 7 = -8$

Let's solve Equation A step by step:

Start with: $-3(x + 7) = 24$

Divide both sides by $-3$: $x + 7 = \frac{24}{-3} = -8$

This gives the same equation as Equation B: $x + 7 = -8$, meaning that the correct statement is d:
"Dividing both sides of Equation A by -3 gives $x + 7 = -8$."

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3. Is 0 a solution to $2x + 10 = 4x + 10$?

We want to check if $x = 0$ satisfies this equation. Substitute $x = 0$ into the equation:

$2(0) + 10 = 4(0) + 10$ $10 = 10$

Since both sides of the equation are equal when $x = 0$, the answer is yes, $0$ is a solution.

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4. Kiran says that a solution to the equation $x + 4 = 20$ must also be a solution to the equation $5(x + 4) = 100$.

To explain this, let’s solve both equations.

First, solve $x + 4 = 20$: $x = 20 - 4 = 16$

Now, substitute $x = 16$ into the second equation: $5(x + 4) = 100$ $5(16 + 4) = 5(20) = 100$

Since $x = 16$ satisfies both equations, Kiran is correct. The solution to $x + 4 = 20$ also works in $5(x + 4) = 100$. This is because multiplying both sides of the first equation by 5 yields the second equation.

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Summary:

• Problem 1: Find the equivalent equation for $7x = 1$.
• Problem 2: The correct answer is d.
• Problem 3: $0$ is indeed a solution.
• Problem 4: The solution to $x + 4 = 20$ must also satisfy $5(x + 4) = 100$ because both equations are related by multiplication.

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Do you want more details or have any questions?
Here are 5 related problems to expand on these:

1. Solve $8x = 2$. What is $x$?
2. Solve the equation $2(x + 5) = 14$.
3. Check if $x = 3$ is a solution to $4x - 5 = 7$.
4. If $3(x - 2) = 12$, what is $x$?
5. Are the equations $x + 2 = 5$ and $2x + 4 = 10$ equivalent?

Tip: Always check solutions by plugging them back into the original equation.