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Common Core: 6th Grade Math : Expressions & Equations

Study concepts, example questions & explanations for Common Core: 6th Grade Math

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All Common Core: 6th Grade Math Resources

6 Diagnostic Tests 186 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #81 : Solve Real World And Mathematical Problems By Writing And Solving Equations: Ccss.Math.Content.6.Ee.B.7

Solve for :

5x=110

Possible Answers:

Correct answer:

Explanation:

The equation 5x=110 asks us to identify which number, , multiplied by is equal to 110 .

In order to solve this equation, we need to isolate the variable on one side of the equals sign. We will do this by performing the opposite of the operations that were done on the variable,. We need to remember that whatever we do to one side of the equals sign, we have to do to the other.

Since was multiplied by the variable, we can divide from both sides of the equation.

$\frac{5x}{5}=\frac{110}{5}$

x=22

We can check to make sure we have the correct value for by plugging into the original equation; thus:

5(22)=110

110=110

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Example Question #82 : Solve Real World And Mathematical Problems By Writing And Solving Equations: Ccss.Math.Content.6.Ee.B.7

Solve for :

4x=68

Possible Answers:

Correct answer:

Explanation:

The equation 4x=68 asks us to identify which number, , multiplied by is equal to .

Since was multiplied by the variable, we can divide from both sides of the equation.

$\frac{4x}{4}=\frac{68}{4}$

x=17

We can check to make sure we have the correct value for by plugging into the original equation; thus:

4(17)=68

68=68

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Example Question #83 : Solve Real World And Mathematical Problems By Writing And Solving Equations: Ccss.Math.Content.6.Ee.B.7

Solve for :

3x=45

Possible Answers:

Correct answer:

Explanation:

The equation 3x=45 asks us to identify which number, , multiplied by is equal to .

Since was multiplied by the variable, we can divide from both sides of the equation.

$\frac{3x}{3}=\frac{45}{3}$

x=15

We can check to make sure we have the correct value for by plugging into the original equation; thus:

3(15)=45

45=45

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Example Question #84 : Solve Real World And Mathematical Problems By Writing And Solving Equations: Ccss.Math.Content.6.Ee.B.7

Solve for :

2x=46

Possible Answers:

Correct answer:

Explanation:

The equation 2x=46 asks us to identify which number, , multiplied by is equal to .

Since was multiplied by the variable, we can divide from both sides of the equation.

$\frac{2x}{2}=\frac{46}{2}$

x=23

We can check to make sure we have the correct value for by plugging into the original equation; thus:

2(23)=46

46=46

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Example Question #91 : Solve Real World And Mathematical Problems By Writing And Solving Equations: Ccss.Math.Content.6.Ee.B.7

Solve for :

15x=105

Possible Answers:

Correct answer:

Explanation:

The equation 15x=105 asks us to identify which number, , multiplied by is equal to 105 .

Since was multiplied by the variable, we can divide from both sides of the equation.

$\frac{15x}{15}=\frac{105}{15}$

x=7

We can check to make sure we have the correct value for by plugging into the original equation; thus:

15(7)=105

105=105

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Example Question #91 : Solve Real World And Mathematical Problems By Writing And Solving Equations: Ccss.Math.Content.6.Ee.B.7

Solve for :

x+5=29

Possible Answers:

Correct answer:

Explanation:

The equation x+5=29 asks us to identify which number, , plus is equal to .

Since was added to the variable, we can subtract from both sides of the equation.

$\frac{\begin{array}[b]{r}x+5=29\\ -5-5\end{array}}{\\\\x+0=24}$

Simplify by dropping the because it has not numerical value; therefore:

x=24

We can check to make sure we have the correct value for by plugging into the original equation; thus:

24+5=29

29=29

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Example Question #1 : Understand Independent And Dependent Variables: Ccss.Math.Content.6.Ee.C.9

Select the table of values that represent the relationship between {m} and {n} if 7 m + 3 = n

Possible Answers:

${\begin{tabular}{|c|c|}\hline m & n \\ \hline 4 & 31\\ \hline 14 & 101\\ \hline 15 & 119\\ \hline 19& 146\\ \hline \end{tabular}}$

${\begin{tabular}{|c|c|}\hline m & n \\ \hline 0 & 3\\ \hline 8 & 118\\ \hline 14 & 202\\ \hline 18& 258\\ \hline \end{tabular}}$

${\begin{tabular}{|c|c|}\hline m & n \\ \hline 0 & 4\\ \hline 8 & 61\\ \hline 14 & 104\\ \hline 18& 133\\ \hline \end{tabular}}$

${\begin{tabular}{|c|c|}\hline m & n \\ \hline 0 & 3\\ \hline 8 & 59\\ \hline 14 & 101\\ \hline 18& 129\\ \hline \end{tabular}}$

Correct answer:

${\begin{tabular}{|c|c|}\hline m & n \\ \hline 0 & 3\\ \hline 8 & 59\\ \hline 14 & 101\\ \hline 18& 129\\ \hline \end{tabular}}$

Explanation:

In the equation 7 m + 3 = n , {m} is the independent variable and {m} is the dependent variable. This means, as we manipulate {m} , {n} will change.

Because we are given tables in our answer choices, we can plug in the given value for {m} from the table and use our equation from the question to see if that equals the value given for {n} in the table.

Let's start by testing values from the following table:

${\begin{tabular}{|c|c|}\hline m & n \\ \hline 0 & 4\\ \hline 8 & 61\\ \hline 14 & 104\\ \hline 18& 133\\ \hline \end{tabular}}$

{7(0)+3=4}

{0+3=4}

${3\neq 4}$

Because this equation did not work out, this means that not all of the values from the table are representative of the relationship between {m} and {n} if {7 m + 3 = n} ; thus, this answer choice is not correct and can be eliminated.

Next, let's test values from the following table:

${\begin{tabular}{|c|c|}\hline m & n \\ \hline 2 & 17\\ \hline 7 & 104\\ \hline 8 & 118\\ \hline 10& 146\\ \hline \end{tabular}}$

{7(2)+3=17}

{14+3=17}

{17=17}

These values are correct for our equation, but to be safe we should plug each value into our equation until we find a value that is not correct, or that each value is correct.

{7(7)+3=104}

{49+3=104}

${52\neq 104}$

${\begin{tabular}{|c|c|}\hline m & n \\ \hline 4 & 31\\ \hline 14 & 101\\ \hline 15 & 119\\ \hline 19& 136\\ \hline \end{tabular}}$

{7(4)+3=31}

{28+3=31}

{31=31}

These values are correct for our equation, but to be safe we should plug each value into our equation until we find a value that is not correct, or that each value is correct.

{7(14)+3=101}

{98+3=101}

{101=101}

Again, these values are correct for our equation, but to be safe we should plug each value into our equation until we find a value that is not correct, or that each value is correct.

{7(15)+3=119}

{105+3=119}

${108\neq 119}$

Finally, let's test values from the following table:

${\begin{tabular}{|c|c|}\hline m & n \\ \hline 11 & 80\\ \hline 12 & 87\\ \hline 16 & 115\\ \hline 17& 122\\ \hline \end{tabular}}$

{7(11)+3=80}

{77+3=80}

{80=80}

These values are correct for our equation, but to be safe we should plug each value into our equation until we find a value that is not correct, or that each value is correct.

{7(12)+3=87}

{84+3=87}

{87=87}

Again, these values are correct for our equation, but to be safe we should plug each value into our equation until we find a value that is not correct, or that each value is correct.

{7(16)+3=115}

{112+3=115}

{115=115}

Again, these values are correct for our equation, but to be safe we should plug each value into our equation until we find a value that is not correct, or that each value is correct.

{7(17)+3=122}

{119+3=122}

{122=122}

All of these values were correct for our equation; thus, this table is our correct answer.

Report an Error

Example Question #2 : Understand Independent And Dependent Variables: Ccss.Math.Content.6.Ee.C.9

Select the table of values that represent the relationship between {m} and {n} if - 13 m + 5 = n

Possible Answers:

${\begin{tabular}{|c|c|}\hline m & n \\ \hline 0 & 5\\ \hline 1 & -16\\ \hline 4 & -94\\ \hline 19& -484\\ \hline \end{tabular}}$

${\begin{tabular}{|c|c|}\hline m & n \\ \hline 6 & -73\\ \hline 10 & -125\\ \hline 17 & -205\\ \hline 18& -219\\ \hline \end{tabular}}$

${\begin{tabular}{|c|c|}\hline m & n \\ \hline 0 & 6\\ \hline 1 & -6\\ \hline 4 & -44\\ \hline 19& -238\\ \hline \end{tabular}}$

${\begin{tabular}{|c|c|}\hline m & n \\ \hline 0 & 5\\ \hline 1 & -8\\ \hline 4 & -47\\ \hline 19& -242\\ \hline \end{tabular}}$

Correct answer:

${\begin{tabular}{|c|c|}\hline m & n \\ \hline 0 & 5\\ \hline 1 & -8\\ \hline 4 & -47\\ \hline 19& -242\\ \hline \end{tabular}}$

Explanation:

In the equation - 13 m + 5 = n , {m} is the independent variable and {m} is the dependent variable. This means, as we manipulate {m} , {n} will change.

Let's start by testing values from the following table:

${\begin{tabular}{|c|c|}\hline m & n \\ \hline 0 & 6\\ \hline 1 & -6\\ \hline 4 & -44\\ \hline 19& -238\\ \hline \end{tabular}}$

{-13(0)+5=6}

{0+5=6}

${5\neq 6}$

Because this equation did not work out, this means that not all of the values from the table are representative of the relationship between {m} and {n} if {- 13 m + 5 = n} ; thus, this answer choice is not correct and can be eliminated.

Next, let's test values from the following table:

${\begin{tabular}{|c|c|}\hline m & n \\ \hline 4 & -47\\ \hline 9 & -224\\ \hline 13 & -328\\ \hline 17& -432\\ \hline \end{tabular}}$

{-13(4)+5=-47}

{-52+5=-47}

{-47=-47}

These values are correct for our equation, but to be safe we should plug each value into our equation until we find a value that is not correct, or that each value is correct.

{-13(9)+5=-224}

{-117+5=-224}

${-112\neq -224}$

${\begin{tabular}{|c|c|}\hline m & n \\ \hline 6 & -73\\ \hline 10 & -125\\ \hline 17 & -205\\ \hline 18& -229\\ \hline \end{tabular}}$

{-13(6)+5=-73}

{-78+5=-73}

{-73=-73}

These values are correct for our equation, but to be safe we should plug each value into our equation until we find a value that is not correct, or that each value is correct.

{-13(10)+5=-125}

{-130+5=-125}

{-125=-125}

Again, these values are correct for our equation, but to be safe we should plug each value into our equation until we find a value that is not correct, or that each value is correct.

{-13(17)+5=-205}

{-221+5=-205}

${-216\neq -205}$

Finally, let's test values from the following table:

${\begin{tabular}{|c|c|}\hline m & n \\ \hline 0 & 5\\ \hline 4 & -47\\ \hline 15 & -190\\ \hline 17& -216\\ \hline \end{tabular}}$

{-13(0)+5=5}

{0+5=5}

{5=5}

These values are correct for our equation, but to be safe we should plug each value into our equation until we find a value that is not correct, or that each value is correct.

{-13(4)+5=-47}

{-52+5=-47}

{-47=-47}

Again, these values are correct for our equation, but to be safe we should plug each value into our equation until we find a value that is not correct, or that each value is correct.

{-13(15)+5=-190}

{-195+5=-190}

{-190=-190}

Again, these values are correct for our equation, but to be safe we should plug each value into our equation until we find a value that is not correct, or that each value is correct.

{-13(17)+5=-216}

{-221+5=-216}

{-216=-216}

All of these values were correct for our equation; thus, this table is our correct answer.

Report an Error

Example Question #3 : Understand Independent And Dependent Variables: Ccss.Math.Content.6.Ee.C.9

Select the table of values that represent the relationship between {m} and {n} if 11 m - 17 = n

Possible Answers:

${\begin{tabular}{|c|c|}\hline m & n \\ \hline 1 & -6\\ \hline 4 & 27\\ \hline 10 & 104\\ \hline 13& 136\\ \hline \end{tabular}}$

${\begin{tabular}{|c|c|}\hline m & n \\ \hline 4 & 27\\ \hline 5 & 38\\ \hline 12 & 115\\ \hline 16& 159\\ \hline \end{tabular}}$

${\begin{tabular}{|c|c|}\hline m & n \\ \hline 4 & 27\\ \hline 5 & 76\\ \hline 12 & 230\\ \hline 16& 318\\ \hline \end{tabular}}$

${\begin{tabular}{|c|c|}\hline m & n \\ \hline 4 & 28\\ \hline 5 & 40\\ \hline 12 & 118\\ \hline 16& 163\\ \hline \end{tabular}}$

Correct answer:

${\begin{tabular}{|c|c|}\hline m & n \\ \hline 4 & 27\\ \hline 5 & 38\\ \hline 12 & 115\\ \hline 16& 159\\ \hline \end{tabular}}$

Explanation:

In the equation 11 m - 17 = n , {m} is the independent variable and {m} is the dependent variable. This means, as we manipulate {m} , {n} will change.

Let's start by testing values from the following table:

${\begin{tabular}{|c|c|}\hline m & n \\ \hline 4 & 28\\ \hline 5 & 40\\ \hline 12 & 118\\ \hline 16& 163\\ \hline \end{tabular}}$

{11(4)+-17=28}

{44+-17=28}

${27\neq 28}$

Because this equation did not work out, this means that not all of the values from the table are representative of the relationship between {m} and {n} if {11 m - 17 = n} ; thus, this answer choice is not correct and can be eliminated.

Next, let's test values from the following table:

${\begin{tabular}{|c|c|}\hline m & n \\ \hline 6 & 49\\ \hline 7 & 120\\ \hline 10 & 186\\ \hline 14& 274\\ \hline \end{tabular}}$

{11(6)-17=49}

{66-17=49}

{49=49}

These values are correct for our equation, but to be safe we should plug each value into our equation until we find a value that is not correct, or that each value is correct.

{11(7)-17=120}

{77-17=120}

${60\neq 120}$

${\begin{tabular}{|c|c|}\hline m & n \\ \hline 1 & -6\\ \hline 4 & 27\\ \hline 10 & 104\\ \hline 13& 136\\ \hline \end{tabular}}$

{11(1)-17=-6}

{11-17=-6}

{-6=-6}

These values are correct for our equation, but to be safe we should plug each value into our equation until we find a value that is not correct, or that each value is correct.

{11(4)-17=27}

{44-17=27}

{27=27}

Again, these values are correct for our equation, but to be safe we should plug each value into our equation until we find a value that is not correct, or that each value is correct.

{11(10)-17=104}

{110-17=104}

${93\neq 104}$

Finally, let's test values from the following table:

${\begin{tabular}{|c|c|}\hline m & n \\ \hline 6 & 49\\ \hline 10 & 93\\ \hline 16 & 159\\ \hline 17& 170\\ \hline \end{tabular}}$

{11(6)-17=49}

{66-17=49}

{49=49}

These values are correct for our equation, but to be safe we should plug each value into our equation until we find a value that is not correct, or that each value is correct.

{11(10)-17=93}

{110-17=93}

{93=93}

Again, these values are correct for our equation, but to be safe we should plug each value into our equation until we find a value that is not correct, or that each value is correct.

{11(16)-17=159}

{176-17=159}

{159=159}

Again, these values are correct for our equation, but to be safe we should plug each value into our equation until we find a value that is not correct, or that each value is correct.

{11(17)-17=170}

{187+-17=170}

{170=170}

All of these values were correct for our equation; thus, this table is our correct answer.

Report an Error

Example Question #1 : Understand Independent And Dependent Variables: Ccss.Math.Content.6.Ee.C.9

Select the table of values that represent the relationship between {m} and {n} if - 10 m - 7 = n

Possible Answers:

${\begin{tabular}{|c|c|}\hline m & n \\ \hline 0 & -7\\ \hline 3 & -74\\ \hline 6 & -134\\ \hline 19& -394\\ \hline \end{tabular}}$

${\begin{tabular}{|c|c|}\hline m & n \\ \hline 0 & -6\\ \hline 3 & -35\\ \hline 6 & -64\\ \hline 19& -193\\ \hline \end{tabular}}$

${\begin{tabular}{|c|c|}\hline m & n \\ \hline 3 & -37\\ \hline 4 & -47\\ \hline 15 & -146\\ \hline 16& -157\\ \hline \end{tabular}}$

${\begin{tabular}{|c|c|}\hline m & n \\ \hline 0 & -7\\ \hline 3 & -37\\ \hline 6 & -67\\ \hline 19& -197\\ \hline \end{tabular}}$

Correct answer:

${\begin{tabular}{|c|c|}\hline m & n \\ \hline 0 & -7\\ \hline 3 & -37\\ \hline 6 & -67\\ \hline 19& -197\\ \hline \end{tabular}}$

Explanation:

In the equation - 10 m - 7 = n , {m} is the independent variable and {m} is the dependent variable. This means, as we manipulate {m} , {n} will change.

Let's start by testing values from the following table:

${\begin{tabular}{|c|c|}\hline m & n \\ \hline 0 & -6\\ \hline 3 & -35\\ \hline 6 & -64\\ \hline 19& -193\\ \hline \end{tabular}}$

{-10(0)+-7=-6}

{0+-7=-6}

${-7\neq -6}$

Because this equation did not work out, this means that not all of the values from the table are representative of the relationship between {m} and {n} if {- 10 m - 7 = n} ; thus, this answer choice is not correct and can be eliminated.

Next, let's test values from the following table:

${\begin{tabular}{|c|c|}\hline m & n \\ \hline 1 & -17\\ \hline 11 & -234\\ \hline 14 & -294\\ \hline 16& -334\\ \hline \end{tabular}}$

{-10(1)-7=-17}

{-10-7=-17}

{-17=-17}

These values are correct for our equation, but to be safe we should plug each value into our equation until we find a value that is not correct, or that each value is correct.

{-10(11)-7=-234}

{-110-7=-234}

${-117\neq -234}$

${\begin{tabular}{|c|c|}\hline m & n \\ \hline 3 & -37\\ \hline 4 & -47\\ \hline 15 & -146\\ \hline 16& -157\\ \hline \end{tabular}}$

{-10(3)-7=-37}

{-30-7=-37}

{-37=-37}

These values are correct for our equation, but to be safe we should plug each value into our equation until we find a value that is not correct, or that each value is correct.

{-10(4)-7=-47}

{-40-7=-47}

{-47=-47}

Again, these values are correct for our equation, but to be safe we should plug each value into our equation until we find a value that is not correct, or that each value is correct.

{-10(15)-7=-146}

{-150-7=-146}

${-157\neq -146}$

Finally, let's test values from the following table:

${\begin{tabular}{|c|c|}\hline m & n \\ \hline 1 & -17\\ \hline 8 & -87\\ \hline 17 & -177\\ \hline 18& -187\\ \hline \end{tabular}}$

{-10(1)-7=-17}

{-10-7=-17}

{-17=-17}

These values are correct for our equation, but to be safe we should plug each value into our equation until we find a value that is not correct, or that each value is correct.

{-10(8)-7=-87}

{-80-7=-87}

{-87=-87}

Again, these values are correct for our equation, but to be safe we should plug each value into our equation until we find a value that is not correct, or that each value is correct.

{-10(17)-7=-177}

{-170-7=-177}

{-177=-177}

Again, these values are correct for our equation, but to be safe we should plug each value into our equation until we find a value that is not correct, or that each value is correct.

{-10(18)-7=-187}

{-180+-7=-187}

{-187=-187}

All of these values were correct for our equation; thus, this table is our correct answer.

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All Common Core: 6th Grade Math Resources

6 Diagnostic Tests 186 Practice Tests Question of the Day Flashcards Learn by Concept

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Common Core: 6th Grade Math : Expressions & Equations

Study concepts, example questions & explanations for Common Core: 6th Grade Math

All Common Core: 6th Grade Math Resources

Example Questions

Example Question #81 : Solve Real World And Mathematical Problems By Writing And Solving Equations: Ccss.Math.Content.6.Ee.B.7

Example Question #82 : Solve Real World And Mathematical Problems By Writing And Solving Equations: Ccss.Math.Content.6.Ee.B.7

Example Question #83 : Solve Real World And Mathematical Problems By Writing And Solving Equations: Ccss.Math.Content.6.Ee.B.7

Example Question #84 : Solve Real World And Mathematical Problems By Writing And Solving Equations: Ccss.Math.Content.6.Ee.B.7

Example Question #91 : Solve Real World And Mathematical Problems By Writing And Solving Equations: Ccss.Math.Content.6.Ee.B.7

Example Question #91 : Solve Real World And Mathematical Problems By Writing And Solving Equations: Ccss.Math.Content.6.Ee.B.7

Example Question #1 : Understand Independent And Dependent Variables: Ccss.Math.Content.6.Ee.C.9

Example Question #2 : Understand Independent And Dependent Variables: Ccss.Math.Content.6.Ee.C.9

Example Question #3 : Understand Independent And Dependent Variables: Ccss.Math.Content.6.Ee.C.9

Example Question #1 : Understand Independent And Dependent Variables: Ccss.Math.Content.6.Ee.C.9

All Common Core: 6th Grade Math Resources

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