Question

eureka math² new york next gen
name
date: dec 15, 2025

1. which equations have infinitely many solutions? choose all that apply.

a 3 = 3
b 5 = -5
c 10a = 10
d. $-\frac{1}{2}m = \frac{1}{2}m$
e. $-6z = -6z$

Explanation:

To determine which equations have infinitely many solutions, we analyze each option:

Step 1: Analyze Option A ($3 = 3$)

This is an identity (always true) with no variables. However, typically, equations with infinitely many solutions involve variables where the equation is true for all values of the variable. Since this has no variable, it's a true statement but not an equation with a variable having infinitely many solutions in the context of variable equations. Wait, actually, if we consider it as an equation (though no variable), but usually, for equations with variables, we look at those where simplifying gives an identity with the variable. But let's re - evaluate. Wait, maybe the initial thought was wrong. Let's check each equation properly.

Step 2: Analyze Option B ($5=-5$)

This is a false statement. So it has no solutions.

Step 3: Analyze Option C ($10a = 10$)

We can solve for $a$ by dividing both sides by 10: $a=\frac{10}{10}=1$. So this equation has a unique solution ($a = 1$), not infinitely many.

Step 4: Analyze Option D ($-\frac{1}{2}m=\frac{1}{2}m$)

Let's add $\frac{1}{2}m$ to both sides: $-\frac{1}{2}m+\frac{1}{2}m=\frac{1}{2}m+\frac{1}{2}m$. This simplifies to $0 = m$. So there is a unique solution ($m = 0$), not infinitely many.

Step 5: Analyze Option E ($-6z=-6z$)

If we add $6z$ to both sides, we get $-6z + 6z=-6z + 6z$, which simplifies to $0 = 0$. This is an identity, which means that no matter what value $z$ takes, the equation will be true. So this equation has infinitely many solutions.

Wait, also, let's re - check option A. If we consider an equation like $3 = 3$ as an equation (even without a variable), it is always true. But in the context of equations with variables, the ones with variables that are identities (like $-6z=-6z$) have infinitely many solutions. Also, if we consider the equation $3 = 3$ as a tautology, but usually, in the context of linear equations with one variable, we look at equations with a variable. But maybe the problem considers equations (including those without variables) in a broader sense. Wait, no, the other options have variables. Let's re - examine:

For an equation of the form $ax + b=cx + d$, if $a = c$ and $b = d$, then it has infinitely many solutions (identity). If $a = c$ and $b
eq d$, no solutions. If $a
eq c$, unique solution.

- Option A: $3 = 3$: No variable, but it's a true statement. However, in the context of the other options (which have variables), maybe the problem expects equations with variables. But let's see the original problem again. The options are:

A. $3 = 3$

B. $5=-5$

C. $10a = 10$

D. $-\frac{1}{2}m=\frac{1}{2}m$

E. $-6z=-6z$

Let's re - analyze each with variables:

- Option E: $-6z=-6z$: Subtract $-6z$ from both sides: $-6z+6z=-6z + 6z\implies0 = 0$. This is an identity, so for all real numbers $z$, the equation holds. So infinitely many solutions.

- Option A: $3 = 3$: No variable. But if we consider it as an equation, it's always true, but there's no variable to have solutions for. So in the context of equations with a variable (since the other options have variables), the correct one with infinitely many solutions is E. Also, is there another? Wait, let's check option A again. If we think of it as an equation in the variable $x$ (even though $x$ is not present), like $3=3$ for all $x$, but that's a stretch. The standard way is that for an equation with a variable, when the coefficients of the variable and the constants on both sides are equal (after simplifying), it's an identity. So $-6z=-6z$ is an identity (coefficients of $z$ are both - 6, and the constants (0 on both sides) are equal). So it has infinitely many solutions. Also, is there a mistake in the initial analysis of option A? Let's see, if we have an equation like $0x+3 = 0x + 3$, which is equivalent to $3 = 3$, then this equation (with variable $x$) has infinitely many solutions because no matter what $x$ is, it's true. So in that case, option A can also be considered. Wait, the equation $3 = 3$ can be written as $0x+3=0x + 3$, which is an identity, so it has infinitely many solutions (for any $x$). So both A and E?

Wait, let's check the original problem's options:

A. $3 = 3$: If we consider it as an equation in $x$ (implicitly), $0x+3 = 0x + 3$, which is true for all $x$, so infinitely many solutions.

E. $-6z=-6z$: As we saw, true for all $z$, so infinitely many solutions.

Wait, what about option D: $-\frac{1}{2}m=\frac{1}{2}m$:

$-\frac{1}{2}m-\frac{1}{2}m=\frac{1}{2}m-\frac{1}{2}m\implies - m=0\implies m = 0$. So unique solution.

Option C: $10a=10\implies a = 1$, unique solution.

Option B: $5=-5$, false, no solution.

So the equations with infinitely many solutions are A and E.

Answer:

A. $3 = 3$, E. $-6z=-6z$