Answer q = 25 Explanation Given: The equation, 9 = 4 + 1 5 q . Concept Used: To get rid of a number in addition from one side, subtract the same number from both sides of equal sign. To get rid of a number in subtraction from one side, add the same number both sides of equal sign. To get rid of a number in multiplication from one side, divide the same number from both sides of equal sign. To get rid of a number in division from one side, multiply the same number both sides of equal sign. Rules of Addition/ Subtraction: Two numbers with similar sign always get added and the resulting number will carry the similar sign. Two numbers with opposite signs always get subtracted and the resulting number will carry the sign of larger number. Rules of Multiplication/ Division: The product/quotient of two similar sign numbers is always positive. The product/quotient of two numbers with opposite signs is always negative. Calculation: In order to solve the given equation 9 = 4 + 1 5 q for the unknown variable, isolate the variable term q on one side by performing some basic algebraic operations to get rid of the other numbers and terms associated with it. Here to isolate q on right side, first subtract 4 from both sides and then multiply both sides by 5 and then simplify further as shown below, 9 = 4 + 1 5 q 9 − 4 = 4 − 4 + 1 5 q 5 = 1 5 q 5 × 5 = 1 5 q × 5 25 = q Thus, the solution of the given equation is q = 25 . Now, to check the solution, substitute q = 25 in the given equation and check whether the values on both sides of equal sign is same or no. So, it gives 9 = 4 + 1 5 q 9 = 4 + 1 5 ( 25 ) 9 = 4 + 5 9 = 9 True Statement Since, the left hand side and right hand side are equal, so the solution is correct.

Glencoe Math Accelerated, Student Edition

1st Edition

ISBN: 9780076637980

Author: McGraw-Hill Glencoe

Publisher: MCG

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Question

Chapter 8.2, Problem 12GP

To determine

The solution of the equation $9=4+15q$ , and check the solution.

Expert Solution & Answer

Answer to Problem 12GP

$q=25$

Explanation of Solution

Given:

The equation, $9=4+15q$ .

Concept Used:

To get rid of a number in addition from one side, subtract the same number from both sides of equal sign.
To get rid of a number in subtraction from one side, add the same number both sides of equal sign.
To get rid of a number in multiplication from one side, divide the same number from both sides of equal sign.
To get rid of a number in division from one side, multiply the same number both sides of equal sign.

Rules of Addition/ Subtraction:

Two numbers with similar sign always get added and the resulting number will carry the similar sign.
Two numbers with opposite signs always get subtracted and the resulting number will carry the sign of larger number.

Rules of Multiplication/ Division:

The product/quotient of two similar sign numbers is always positive.
The product/quotient of two numbers with opposite signs is always negative.

Calculation:

In order to solve the given equation $9=4+15q$ for the unknown variable, isolate the variable term q on one side by performing some basic algebraic operations to get rid of the other numbers and terms associated with it.

Here to isolate q on right side, first subtract 4 from both sides and then multiply both sides by 5 and then simplify further as shown below,

$9=4+15q9−4=4−4+15q5=15q5×5=15q×525=q$

Thus, the solution of the given equation is $q=25$ .

Now, to check the solution, substitute $q=25$ in the given equation and check whether the values on both sides of equal sign is same or no. So, it gives

$9=4+15q9=4+15(25)9=4+59=9 True Statement$

Since, the left hand side and right hand side are equal, so the solution is correct.

Chapter 8.2, Problem 11GP

Chapter 8.2, Problem 13GP

Chapter 8 Solutions

Glencoe Math Accelerated, Student Edition

Ch. 8.1 - Prob. 1GP Ch. 8.1 - Prob. 2GP Ch. 8.1 - Prob. 3GP Ch. 8.1 - Prob. 4GP Ch. 8.1 - Prob. 5GP Ch. 8.1 - Prob. 6GP Ch. 8.1 - Prob. 7GP Ch. 8.1 - Prob. 8GP Ch. 8.1 - Prob. 9GP Ch. 8.1 - Prob. 10GP

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The solution of the equation 9 = 4 + 1 5 q , and check the solution.