Monomials, Polynomials, Linear and Quadratic Equations, System of Equations

Question 1

What is the definition of an equality?

Answer 1

An equality is a mathematical statement that two expressions are equal.

The value of the left side equates to the value of the right side.

Example:

3 + 3 = 6

2x + 5x = 7x

Question 2

What is the definition of an equation?

Answer 2

An equation is an equality in which the unknown has a specific value.

Example:

2x = 8

In this case x can only one value, 4. 4 is a solution of the equation above.

Question 3

What equations do we call equivalent?

Answer 3

Equations that have the same solution are called equivalent equations.

In some cases, you may need to transform the original equation into an equivalent to solve it.

Example:

2x + 4 = 8 and 2x + 4 - 8 = 0

Question 4

Define:

• Variable
• Constant
• Coefficient

Answer 4

• A variable is a symbol for an unknown number
• A coefficient is a multiplicative factor of a variable
• Constant is a fixed value

Example:

5x - 2 = 8

5 is a coefficient, x is a variable, 2 and 8 are constants.

Question 5

How do you solve an equation with one variable?

Answer 5

• Simplify each side of the equation if necessary
• Isolate the variable by using inverse operations

The result will be an equation with a variable on one side and a real number on the other.

Example:

6x - 8 = -3x + 10

Get all the x’s on one side by adding 3x to both sides: 9x - 8 = 10. Then, add 8 to both sides: 9x = 18. Then, divide both sides by 9: x = 2

Question 6

In order to solve a linear equation, you might need to simplify it. What operations might be necessary?

Answer 6

You may need to do all or some of these to simplify the original equation:

1. Combine similar terms within grouping symbols
2. Use distributive property
3. Remove unnecessary parentheses
4. Combine like terms

Once you’ve simplified the original equation, proceed with isolating the variable by using inverse operations.

Example:

4 (3x + 3 - 8x) - (-6x) = 26

1. 4 (3 - 5x) - (-6x) = 26
2. 12 - 20x - (-6x) = 26
3. 12 - 20x + 6x = 26
4. 12 - 14x = 26

Question 7

What are inverse operations?

Answer 7

Two operations are inverse to each other when one operation reverses the effect of the other operation.

Addition and subtraction are inverse operations. Multiplication and division are inverse operations.

Question 8

How would you use inverse operations to solve the following equations?

z + 15 = 32

5x = 65

Answer 8

z + 15 = 32

15 is added to z. To find z, subtract 15 from 32.

5x = 65

5 and x are multiplied. To find x, divide 65 by 5.

Question 9

What does it mean “to isolate the variable” in an equation?

Answer 9

Isolating the variable is the process the result of which is an equation with the variable on one side and a real number on the other side.

To isolate the variable, use inverse operations.

Example:

12 - 14x = 26

• 14x = 26 - 12
• 14x = 14
• x* = -1

Question 10

Formulate the rule of equality for solving equations.

Answer 10

In plain language, whatever you do to one side of the equation, you must do to the other.

The rule of equality states that the same operation using equal numbers must be done on both sides of an equation.

• Example:*
• x* + 4 = 6
• x* + 4 + 3 = 6 + 3

If for some reason you need to add 3 to the left side of the equation, you must also add 3 to the right side.

Question 11

What steps might be necessary to solve an inequality?

Answer 11

Solve an inequality the same way you would solve an equation, i.e. simplify and isolate the variable.

*** Note: When multiplying or dividing both side of the inequality by the same negative number, it reverses the direction of the sign of the inequality.

Question 12

How does multiplying or dividing an inequality by a negative number affect the direction of the sign of the inequality?

Answer 12

Multiplying or dividing an inequality by a negative number reverses the direction of the inequality.

Example:

-2x > 10 Dividing both sides by -2

x -5

Question 13

Define:

A monomial.

Answer 13

A monomial is an expression that has one term.

It’s either a real number, a variable, or the product of a real number and one or more variables.

*** A variable cannot be in the denominator.

Example:

5x, 3a²b, 10

Question 14

How do you figure out the degree of a monomial?

Answer 14

The degree of a monomial is the sum of the degrees (exponents) of its variables.

Example:

x⁵y³z

5 + 3 + 1 = 9

9 is the degree of this monomial.

Question 15

Define:

A polynomial.

Answer 15

A polynomial is the sum or the difference of monomials.

*** A polynomial must have the same variable.

Example:

5x² + 6x - 7 or

x - 4

Question 16

What are the terms of a polynomial?

Answer 16

Monomials that make up a polynomial are called its terms.

*** Terms are separated by addition or subtraction signs, but never by multiplication signs.

Example:

15 + 10x

15 and 10x are the terms of this polynomial.

Question 17

How many terms do binomials and trinomials have?

Answer 17

• Binomial is a polynomial with two terms
• Trinomial is a polynomial with three terms

Question 18

How do you simplify a polynomial?

Answer 18

To simplify a polynomial, combine similar terms.

Similar terms contain the same variables and same exponents.

Example:

3 + 5x² + 10x + 24x² - 3x + 7 =

10 + 7x + 29x²

Question 19

How do you figure out the degree of a polynomial?

Answer 19

The degree of a polynomial is the highest degree of any of its terms.

Example:

7x⁴ + 10x

Evaluate both terms. The exponent of the 1st term is 4; the exponent of the 2nd term is 1. Therefore, the degree of this binomial is 4.

Question 20

How do you add and subtract monomials?

Answer 20

Monomials being added or subtracted must have the same variable(s).

They are called like terms.

Example:

10x + 2x = 12x

10x and 2x are like terms. To add them, just add the coefficients.

Question 21

How do you multiply monomials?

Answer 21

Multiply the coefficients and the variables of the monomials separately. Write the product in the exponential form when multiplying same bases.

Example:

8a x 2a = (8 x 2) (a x a) = 16a²

Question 22

How do you add or subtract polynomials?

Answer 22

Combine the terms with exactly the same variable(s); i.e. like terms.

Example:

4x - 7 + 5x + 3x² + 3 = 3x² + 9x - 4

4x and 5x are like terms. So are -7 and 3.

Question 23

How do you multiply a monomial by a polynomial?

a (b + c + d)

Answer 23

To multiply a polynomial by a monomial, use the distributive property.

a(b + c + d) = ab + ac + ad

Question 24

What acronym helps you remember how to multiply two binomials?

(a + b)(c + d)

Answer 24

To multiply two binomials, use the FOIL method.

• (a + b)(c + d) =*
• ac + ad + bc + bd*

FOIL is an acronym for:

F - product of the FIRST terms

O - product of the OUTERMOST terms

I - product of the INNERMOST terms

L - product of the LAST terms

Question 25

How do you multiply polynomials with *more* than *two* terms? *(a + b)(c + d + e)*

Answer 25

To multiply two polynomials with *more* than *two* terms, multiply *each* term of the first polynomial by *each* term of the second. Simply use the distributive property, then combine like terms. ***(a + b) (c + d + e) =*** ***ac + ad + ae + bc + bd + be*** *Example:* (*x* + 3)(*x*² + 5*x* + 6) = (*x* + 3)*x*² + (*x* + 3)5*x* + (*x* + 3)6 = * x*³ + 3*x*² + 5*x*² + 15*x* + 6*x* + 18 = * x*³ + 8*x*² + 21*x* + 18

Question 26

What does the product of the sum and the difference of two monomials equal to? (*a + b*)(*a - b*) = ?

Answer 26

(*a - b*)(*a + b*) *= **a² - b²*** The product of the sum and the difference of two monomials equals to the **difference** of their **squares**.

Question 27

**(*a + b*)²** = ? **(*a - b*)²** = ?

Answer 27

***(a + b)² **=* ***a²+ 2ab + b²*** ***(a - b)²** =* ***a²- 2ab + b²***

Question 28

**(*a + b*)³**= ? **(*a - b*)³** = ?

Answer 28

Below are the formulas for the cube of the sum or the difference of two monomials. ***(a + b)³* =** ***a³ + 3a²b + 3a**b² + b³*** ***(a - b)³**=*** ***a³ - 3a²b + 3ab² - b³ ***

Question 29

Find the greatest common factor (GCF) of the expressions below. ***ca + cb* = ?** ***ca - cb* = ?**

Answer 29

Factor out *c* as shown below: ***ca + cb = c(a + b)*** ***ca - cb = c(a - b)***

Question 30

What is a *prime* polynomial?

Answer 30

A *prime* polynomial is a polynomial that *cannot* be factored.

Question 31

Is there a way to factor out the sum of squares of two monomials? *a² + b²*

Answer 31

The sum of squares *a²+ b² *is a *prime* polynomial. It *cannot* be factored out.

Question 32

How do you factor out the following expression? ***a*² - *b*²**

Answer 32

**Difference of squares** formula: ***a*² - *b*² = (*a + b*)(*a - b*)** *Example*: 36 - 4*x*² = 6² - 2²*x*² = (6 + 2*x*) (6 - 2*x*)

Question 33

How do you factor *perfect square* trinomials? * a² + 2ab + b²* * a² - 2ab + b²*

Answer 33

Factor *perfect square* trinomials as shown below: ***a² + 2ab + b²** =* (a + b)(a + b) *= **(a** **+ b)²*** ***a² - 2ab + b²** =* (a - b)( a - b) *= **(a - b)²***

Question 34

How do you factor the sum or the difference of cubes? * a³ + b³* * a³ - b³*

Answer 34

Factor the sum and the difference of cubes as shown below: * a³ **-** b³ = (a **-** b)(a² **+** ab + b²)* * a³ + b³ = (a **+** b)(a² **-** ab + b²* \*\*\* Watch the signs. Use SOAP for help memorize it. "Same, Opposite, Always Positive".

Question 35

What kind of a polynomial is a **quadratic** equation? What is the general form of a quadratic equation?

Answer 35

A **quadratic** equation is a polynomial equation of a *second* degree. The general form is *ax² + bx + c = 0* where * a* - quadratic coefficient, * b* - linear coefficient, * c* - constant term, * x* - variable.

Question 36

How do you factor a second degree polynomial into prime polynomials?

Answer 36

* Look for *GMF* (Greatest Monomial Factor) * In binomials, look to apply the formula for the difference of squares * In trinomials, look for a perfect square or a pair of binomial factors \*\*\* Remember that NOT all polynomials can be factored.

Question 37

At most, how many solutions does a quadratic equation have?

Answer 37

At most, any quadratic equation has two solutions. They are also called **roots** of the equation. Some equations may have only one solution or no real solutions.

Question 38

What methods do you know for solving a quadratic equation?

Answer 38

There are four methods to solving quadratic equations: * Factoring * Using square root property * Completing the square * Using quadratic formula \*\*\* The most used methods for the SAT problems are factoring and using square root property.

Question 39

Out of four methods (factoring, completing the square, using the quadratic formula or using the square root property), how do you know which method to use to solve a quadratic equation at hand?

Answer 39

* *Factoring* can be used if * ax² + bx + c* can be factored * Use the *square root* property method if * x² = c* or *x² + b = c* where c is not zero * The *quadratic formula* and *completing the square* methods can be used for any quadratic equation

Question 40

How do you use factoring method to solve a quadratic equation? *ax² + bx + c* = 0

Answer 40

* If possible, factor the left side of the equation * Make each factor equal to zero and solve two linear equations * Example:* * x*² + 15 = 8*x* ⇒ *x*² - 8*x* + 15 = 0 (*x* - 3)(*x* - 5) = 0 * x* - 3 = 0 or *x* - 5 = 0 * x* = 3 or *x* = 5

Question 41

How do you factor a quadratic equation when the *leading coefficient a = 1*? *ax² + bx + c* = 0 \*\*\* This type of quadratic equations is most commonly used on the SAT. Become an expert in simple factoring.

Answer 41

Here is the teorem of the sum and the product of the roots in a quadratic equation: * x² + bx + c* = 0 * The *sum* of the roots is the *negative* of *b* coefficient: root 1 + root 2 = **-***b* * The *product* of the roots is the constant term *c*: root 1 \* root 2 *= c*

Question 42

Find the roots of the following quadratic equation. *x² +* 5*x +* 6 = 0

Answer 42

*x² +* 5*x +* 6 = 0 ## Footnote The *sum* of the roots equals -5. The *product* of the roots equals 6. You need to find two numbers that satisfy both conditions. Since the product has to be positive, the numbers have to be the same sign. But since the sum has to be negative, you know that these numbers are negative. Root 1 = -2 Root 2 = -3

Question 43

How do you factor the following quadratic equation? *x² +* 5*x +* 6 = 0

Answer 43

*x² +* 5*x +* 6 = 0 The *sum* of the roots equals -5. The *product* of the roots equals 6. Root 1 = -2 Root 2 = -3 Now, let's factor the equation. R₁ = -2 ⇒ *x* + 2 = 0 R₂ = -3 ⇒ *x* + 3 = 0 (*x* + 2)(*x* + 3) = *x*² + 5*x* + 6

Question 44

If both *c* and *b* in a quadratic equation are *positive*, what can you conclude about the roots of this equation? *x² + bx + c* = 0

Answer 44

The roots of this equation are both *negative*. Given that *c* is positive, the product of the roots must be positive and therefore, the roots must be the same sign. root 1 \* root 2 = *c* (-root 1) \* (-root 2) = *c* Since the sum of the roots equals to the *negative* of *b*, you can conclude the roots of the equation are both negative. root 1 + root 2 = -*b* * Example:* * x*² + 7*x* + 12 = 0 root 1 = -3 root 2 = -4

Question 45

If *c* is *positive* and *b* is *negative* in a quadratic equation, what can you conclude about the roots of this equation? *x² - bx + c* = 0

Answer 45

The roots of this equation are both *positive*. Given than *c* is positive, the product of the roots is positive. Therefore, the roots must have the same sign. root 1 \* root 2 = *c* ( -root 1) \* ( -root 2) = *c* The sum of the roots equals to the negative of b. Negative of a negative is positive. Two negative numbers cannot add up to a positive sum, therefore, the roots must be positive. root 1 + root 2 = - (-*b*) = *b* * Example:* * x*² - 7*x* + 12 = 0 root 1 = 3 root 2 = 4

Question 46

If *c* is *negative* and *b* is *negative* in a quadratic equation, what can you conclude about the roots of this equation? *x² - bx - c* = 0

Answer 46

The roots are of opposite signs and the larger root is *positive*. Given than *c* is *negative*, the product of the roots is negative. Therefore, the roots must be of opposite signs. root 1 \* ( -root 2) = *-c* (-root 1) \* root 2 = -*c* The sum of the roots equals to the *negative* of b. Negative of a negative is positive. The *larger* root of the equation must be *positive*. root 1 + root 2 = - (-*b*) = *b* * Example:* * x*² - 5*x* - 6 = 0 root 1 = 6 root 2 = - 1

Question 47

If *c* is *negative* and *b* is *positive* in a quadratic equation, what can you conclude about the roots of this equation? *x² + bx - c* = 0

Answer 47

The roots of this equation are of opposite signs and the larger root is *negative*. Given than *c* is *negative*, the product of the roots is negative. Therefore, the roots must be of opposite signs. root 1 \* ( -root 2) = *-c* (-root 1) \* root 2 = -*c* The sum of the roots equals to the *negative* of b. The *larger* root of the equation must be *negative*. root 1 + root 2 = -*b* * Example:* * x*² + 5*x* - 6 = 0 root 1 = -6 root 2 = 1

Question 48

How do you factor a quadratic equation when the leading coefficient *a* is *not* 1? * ax² + bx + c* * a, b, c* are real numbers and *a* is not zero and not 1.

Answer 48

The theorem states that: * The *sum* of the roots equals the *negative* of the quotient of *b* and *a*: root1 + root 2 *= **-** ^b/_a* * The *product* of the roots equals the quotient of *c* and *a*: root1 \* root2 = *^c/_a*

Question 49

Find the roots of the following quadratic equation? *2x² + x -* 6 = 0

Answer 49

2*x*² + *x* - 6 = 0 The *sum* of the roots equals the *negative* of the quotient of *b* and *a*: root1 + root 2 *= **-** ^b/_a* = -¹*/*₂ The *product* of the roots equals the quotient of *c* and *a*: root1 \* root2 = *^c/_a* = -⁶*/*₂ = -3 The numbers that satisfy both conditions above are -2 and ³/_2.

Question 50

How do you use the *square root* method to solve a quadratic equation?

Answer 50

Isolate the squared variable on one side and have the constant on the other side. Then, take the *square* *root* of each side of the equation. ## Footnote \*\*\* Remember, the square root of a number yields both a positive and a negative value. *Example:* 3*x*² - 9 = 3 ⇒ 3*x*² = 12 *x*² = 4 ⇒ *x* = 2 or -2

Question 51

When should you use the square root method for solving a quadratic equation?

Answer 51

The square root method is best to use when the equation *only* contains x² term. * ax² = c* * Example:* (*x* - 4)² = 9 - take the square root of each side of the equation. * x* - 4 = 3 ⇒ *x* = 7 * x* - 4 = -3 ⇒ *x* = 1

Question 52

Write the **quadratic** **formula.**

Answer 52

You can find solutions to any ax² + bx + c quadratic equation by using the quadratic formula:

Question 53

When do you use the **quadratic formula** to solve a quadratic equation?

Answer 53

You can use the quadratic formula on any and all quadratics in the form of

ax² + bx + c

where a is not zero.

Plug in the values of a, b and c into the formula.

Question 54

Write the formula used to calculate the **discriminant** in a quadratic equation.

Answer 54

In a quadratic equation in the form of *ax² + bx + c* the **discriminant** is calculated by using the formula below: *D = b² - 4ac*

Question 55

How does the discriminant help determine the solutions for a quadratic equation?

Answer 55

***D = b² - 4ac*** * If D \> 0, the equation has *two* real roots * If D = 0, the equation has *one* real root * If D no real roots The discriminant determines the *number* and the kinds of solutions to a quadratic equation in the form of ax2 + bx + c.

Question 56

How do you use **completing the square** method to solve a quadratic equation?

Answer 56

*ax² + bx + c* = 0 * Transpose *c* to the right side * Add a *square* of *half of b* coefficient to *both* sides to make the left side a perfect square *Example*: x² + 6x + 2 = 0 - transpose x² + 6x = -2 - add the square of half of 6 x² + 6x + 9 = -2 + 9 (x + 3)² = 7 - use the square root method *x* = −3 ± ![]()

Question 57

When do you use **completing the square** method to solve a quadratic equation?

Answer 57

*ax² + bx + c* = 0 This method works when *a* equals 1. If *a* is not 1, divide all terms of the equation by *a*. \*\*\* This method is not commonly used to solve equations on the SAT. You will either factor a quadratic or use the quadratic formula to find the roots on the test.

Question 58

What do we call a **system of linear equations**?

Answer 58

When you have *more than one* linear equation with the *same* set of unknowns, it's called a **system** **of linear equations**. The solutions must satisfy every equation in the system. *Example*: 2*x* + 5*y* = 10 -5*y* + 3*x* = 15

Question 59

What methods of solving a system of linear equations do you know?

Answer 59

You can solve a system of linear equations using one of these methods: * graphing * substitution * addition or subtraction The simplest kind of a linear system involves two equations and two variables.

Question 60

How do you solve a system of linear equations with two variables using the graphing method?

Answer 60

Simply draw the graph of each equation on the same coordinate plane. * If the lines *intersect*, the coordinates of the point of the intersection are the solution * If the lines are *parallel*, there is no solution * If the lines are *identical* (coinside), each point on the line is a solution

Question 61

How do you solve a system of two linear equations with two variables using the substitution method?

Answer 61

* Isolate the variable with the coefficient 1 in one equation * Substitute the same variable with the resulted expression in the equation you haven't used. Solve the equation * Use the solution to substitute the unknown in the first equation and solve *Example:* *x* + 2*y* = 18 2*x* + 7*y* = 6 *x* = 18 - 2*y* - isolate the variable 2(18 - 2*y*) + 7*y* = 6 - substitute *x* = 18 - 2*y* 36 - 4*y* + 7*y* = 6 ⇒ 3*y* = -30 ⇒ ***y* = -10** *x* + 2 (-10) = 18 ⇒ ***x* = 38** - plug in *y*

Question 62

When should you use the substitution method to solve a system of linear equations?

Answer 62

Use the substitution method when the coefficient of one of the variables equals 1 or -1. ## Footnote *Example:* *x* + 5*y* = 12 3*x* - 2*y* = 15 *x* = 12 - 5*y* 3(12 - 5*y*) - 2*y* = 15 Solve the second equation for *y*. Then, plug in *y* value into the first equation and solve for *x*.

Question 63

How do you solve a system of two linear equations with two variables using the addition or subtraction method?

Answer 63

* Add or subtract equations to eliminate one variable * Solve the equation resulting from it * Use the solution to substitute the unknown in either equation and solve *Example:* 5*x* + 3*y* = 25 *x* - 3*y* = 11 Add the equations to eliminate *y* variable. 6*x* = 36 ⇒ *x* = 6 - plug in the *x* value and solve for *y.*

Question 64

When should you use the addition or subtraction method in a system of linear equations?

Answer 64

Use this method if the coefficients of one of the variables have the same absolute value. ## Footnote \*\*\* To eliminate the variable, *subtract* one equation from the other if the coefficients are the *same*. *Add* the equations if the coefficients are *opposite*.

Question 65

Sometimes extra steps are necessary before you can use the addition or subtraction method to solve a system of linear equations.

Answer 65

Sometimes you need to *rearrange* the terms in the equations prior to solving them. *Example:* 2(*x* - 3) = -6*y* + 5 5(*y* + 2) = *x* - 3 2*x* - 6 = -6*y* + 5 5*y* + 10 = *x* - 3 2*x* + 6*y* = 11 5*y* - *x* = -13 Now you can solve the system of equations above by using the substitution method.

Question 66

You may need to use multiplication with the addition or subtraction method to solve a system of linear equations.

Answer 66

You may need to *multiply* the equations to make the coefficients the same absolute value. ## Footnote *Example:* Evaluate the GCM of the like terms' coefficients. It's easier to work with 6 than with 20. Multiply the first equation by 2. Multiply the second equation by 3. Then, use the subtraction method. 3*x* - 5*y* = 19 2*x* - 4*y* = 16 6*x* - 10*y* = 38 6*x* - 12*y* = 48 2*y* = -10 ⇒ ***y* = -5** 2*x* - 4 (-5) = 16 ⇒ ***x* = -2**

Question 67

When should you multiply equations prior to using the addition or subtraction method to solve a system of linear equations?

Answer 67

Multiply if: * The coefficient of a variable is a factor of the coefficient of the same variable in another equation * The coefficients are relatively prime or fractional Multiply one or both equations so that the coefficients of one of the variable will have the same absolute value.

Question 68

How many real numbers satisfy this equation? 2*x*² + 4 = 0

Answer 68

There are no real numbers that satisfy the equation 2*x*² + 4 = 0. 2*x*² = -4 *x*² = -2 No square root can be taken of a negative number within a system of real numbers.

Question 69

The squares of two integers are equal. The product of these integers *could* be (a) -3,000 (b) 5,000 (c) 6,000 (d) -9,000

Answer 69

(d) -9,000 ## Footnote If the squares of two integers are the same, the numbers could be opposite like 30 and -30.

Question 70

The product of the roots of (*x* - 15)(*x* + 11) = 0 is (a) - 165 (b) 26 (c) 155 (d) 165

Answer 70

(a) -165 ## Footnote The roots of this equation are 15 and -11. The product must be a negative number.

Question 71

If *a + b* = 5 and *a - b* = 15, then *a² - b²* equals ## Footnote (a) 10 (b) 20 (c) 55 (d) 75

Answer 71

(d) 75 The difference of squares formula states that ***a² - b² = (a + b)(a - b)*** = 5 x 15 = 75

Question 72

(*a +* 2) divided by (4 *-* *a²)* equals ## Footnote (a) *a* + 2 (b) 2 - *a* (c) *a* - 2 (d) 2*a*

Answer 72

(b) 2 - *a* Use the difference of squares formula to factor 4 - *a*². ***a² - b² = (a + b)(a - b)*** 2² - *a*² = (2 + *a*)(2 - *a*) Then, reduce the resulting algebraic fraction by eliminating common factors.

Question 73

How do you *evaluate* an *algebraic expression*?

Answer 73

If you are given the values of the variables, you can *evaluate* an *algebraic expression* by plugging those values in and calculating the result. *Example:* Evaluate the algebraic expression 6 - 3*x* + 28 when *x* = -3. Combine the like terms 6 and 28 and plug in (-3) into the expression (watch the signs). 34 - 3 (-3) = 34 + 9 = 43

Question 74

When you asked to *express* one variable *in terms* of other variables, what do you do?

Answer 74

To "*solve or express in terms*" means to isolate one variable on one side of the equation, leaving the other variables on the other side of the equation. ## Footnote Example:

Question 75

Changing words into algebraic expressions is necessary to solve word problems. What are some of the words you can name that mean the addition operation?

Answer 75

Here are some words that translate into addition: * combine * gain * exceeds * grow * rise * in all * increased by * greater than * more than * larger than * sum * total

Question 76

Changing words into algebraic expressions is necessary to solve word problems. What are some of the words you can name that mean the subtraction operation?

Answer 76

Here are some words that translate into subtraction: * remove * deduct * depriciate * shorten * drop * lose * lower * left * difference * decreased by * fewer * less than * smaller than * take away

Question 77

Changing words into algebraic expressions is necessary to solve word problems. What are some of the words you can name that mean multiplication operation?

Answer 77

Here are some words that translate into multiplication: * double * twice * triple * quadruple * times * squared * cubed * factor * product * multiple of

Question 78

Changing words into algebraic expressions is necessary to solve word problems. What are some of the words you can name that mean division operation?

Answer 78

Here are some words that translate into division: * ratio * average * split * half * per * quotient * third * fourth * part of