Describe Situations in Which Opposite Quantities Combine to Make 0: CCSS.Math.Content.7.NS.A.1a - 7th Grade Math

Card 0 of 148

Question

Compute the following:

Answer

Convert all the double signs to a single sign before solving. Remember, two minus (negative) signs combine to form a plus (positive) sign, and a plus (positive) sign and a minus (negative) sign combine to form a minus (negative) sign.

Compare your answer with the correct one above

Question

Which value, when substituted for x, will make the equation true?

Answer

In order to answer this question, we can solve for . When solving for we need to isolate the variable on one side of the equation.

We can add to both sides in order to isolate the variable, .

$\frac{\begin{array}[b]{r}15-x=0\ +x+x\end{array}}{\\15=x}$

Compare your answer with the correct one above

Question

For the equation provided, what value, when substituted for , will equal

Answer

In order to answer this question, we can solve for . When solving for we need to isolate the variable on one side of the equation.

We can add to both sides in order to isolate the variable, .

$\frac{\begin{array}[b]{r}7-x=0\ +x+x\end{array}}{\\7=x}$

Compare your answer with the correct one above

Question

For the equation provided, what value, when substituted for , will equal

Answer

In order to answer this question, we can solve for . When solving for we need to isolate the variable on one side of the equation.

We can add to both sides in order to isolate the variable, .

$\frac{\begin{array}[b]{r}13-x=0\ +x+x\end{array}}{\\13=x}$

Compare your answer with the correct one above

Question

For the equation provided, what value, when substituted for , will equal

Answer

In order to answer this question, we can solve for . When solving for we need to isolate the variable on one side of the equation.

We can add to both sides in order to isolate the variable, .

$\frac{\begin{array}[b]{r}27-x=0\ +x+x\end{array}}{\\27=x}$

Compare your answer with the correct one above

Question

For the equation provided, what value, when substituted for , will equal

Answer

In order to answer this question, we can solve for . When solving for we need to isolate the variable on one side of the equation.

We can add to both sides in order to isolate the variable, .

$\frac{\begin{array}[b]{r}34-x=0\ +x+x\end{array}}{\\34=x}$

Compare your answer with the correct one above

Question

For the equation provided, what value, when substituted for , will equal

Answer

In order to answer this question, we can solve for . When solving for we need to isolate the variable on one side of the equation.

We can add to both sides in order to isolate the variable, .

$\frac{\begin{array}[b]{r}6-x=0\ +x+x\end{array}}{\\6=x}$

Compare your answer with the correct one above

Question

For the equation provided, what value, when substituted for , will equal

Answer

In order to answer this question, we can solve for . When solving for we need to isolate the variable on one side of the equation.

We can add to both sides in order to isolate the variable, .

$\frac{\begin{array}[b]{r}94-x=0\ +x+x\end{array}}{\\94=x}$

Compare your answer with the correct one above

Question

For the equation provided, what value, when substituted for , will equal

Answer

In order to answer this question, we can solve for . When solving for we need to isolate the variable on one side of the equation.

We can add to both sides in order to isolate the variable, .

$\frac{\begin{array}[b]{r}83-x=0\ +x+x\end{array}}{\\83=x}$

Compare your answer with the correct one above

Question

For the equation provided, what value, when substituted for , will equal

Answer

In order to answer this question, we can solve for . When solving for we need to isolate the variable on one side of the equation.

We can add to both sides in order to isolate the variable, .

$\frac{\begin{array}[b]{r}75-x=0\ +x+x\end{array}}{\\75=x}$

Compare your answer with the correct one above

Question

For the equation provided, what value, when substituted for , will equal

Answer

In order to answer this question, we can solve for . When solving for we need to isolate the variable on one side of the equation.

We can add to both sides in order to isolate the variable, .

$\frac{\begin{array}[b]{r}66-x=0\ +x+x\end{array}}{\\66=x}$

Compare your answer with the correct one above

Question

For the equation provided, what value, when substituted for , will equal

Answer

In order to answer this question, we can solve for . When solving for we need to isolate the variable on one side of the equation.

We can add to both sides in order to isolate the variable, .

$\frac{\begin{array}[b]{r}59-x=0\ +x+x\end{array}}{\\59=x}$

Compare your answer with the correct one above

Question

For the equation provided, what value, when substituted for , will equal

Answer

In order to answer this question, we can solve for . When solving for we need to isolate the variable on one side of the equation.

We can add to both sides in order to isolate the variable, .

$\frac{\begin{array}[b]{r}46-x=0\ +x+x\end{array}}{\\46=x}$

Compare your answer with the correct one above

Question

For the equation provided, what value when substituted for , will equal

Answer

In order to answer this question, we can solve for . When solving for we need to isolate the variable on one side of the equation.

We can subtract to both sides in order to isolate the variable, .

$\frac{\begin{array}[b]{r}11+x=0\ -11\ \ \ \ -11\end{array}}{\\x=-11}$

Compare your answer with the correct one above

Question

For the equation provided, what value when substituted for , will equal

Answer

In order to answer this question, we can solve for . When solving for we need to isolate the variable on one side of the equation.

We can subtract to both sides in order to isolate the variable, .

$\frac{\begin{array}[b]{r}24+x=0\ -24\ \ \ \ -24\end{array}}{\\x=-24}$

Compare your answer with the correct one above

Question

For the equation provided, what value when substituted for , will equal

Answer

In order to answer this question, we can solve for . When solving for we need to isolate the variable on one side of the equation.

We can subtract to both sides in order to isolate the variable, .

$\frac{\begin{array}[b]{r}38+x=0\ -38\ \ \ \ -38\end{array}}{\\x=-38}$

Compare your answer with the correct one above

Question

For the equation provided, what value, when substituted for , will equal

Answer

In order to answer this question, we can solve for . When solving for we need to isolate the variable on one side of the equation.

We can subtract from both sides in order to isolate the variable, .

$\frac{\begin{array}[b]{r}40+x=0\ -40\ \ \ \ -40\end{array}}{\\x=-40}$

Compare your answer with the correct one above

Question

For the equation provided, what value when substituted for , will equal

Answer

In order to answer this question, we can solve for . When solving for we need to isolate the variable on one side of the equation.

We can subtract to both sides in order to isolate the variable, .

$\frac{\begin{array}[b]{r}59+x=0\ -59\ \ \ \ -59\end{array}}{\\x=-59}$

Compare your answer with the correct one above

Question

For the equation provided, what value when substituted for , will equal

Answer

In order to answer this question, we can solve for . When solving for we need to isolate the variable on one side of the equation.

We can subtract to both sides in order to isolate the variable, .

$\frac{\begin{array}[b]{r}62+x=0\ -62\ \ \ \ -62\end{array}}{\\x=-62}$

Compare your answer with the correct one above

Question

For the equation provided, what value when substituted for , will equal

Answer

In order to answer this question, we can solve for . When solving for we need to isolate the variable on one side of the equation.

We can subtract to both sides in order to isolate the variable, .

$\frac{\begin{array}[b]{r}70+x=0\ -70\ \ \ \ -70\end{array}}{\\x=-70}$

Compare your answer with the correct one above

Tap the card to reveal the answer