Solve: (a) 2t = 80 (b) -6.02 = -8.6t. We will use a property of equality to isolate the variable on one side of the equation. To solve the original equation, we want to find a simpler equivalent equation of the form t = a number or a number = t, whose solution is obvious. (a) To isolatet on the left side, we use the division property of equality. We can undo the multiplication by 2 by dividing both sic 2t = 80 This is the equation to solve. 2t 80 Use the division property of equality: Divide both sides by 2. %3D 2 it = Simplify 4 by removing the common factor of 2 in the numerator and denominator: = 1. t = 40 The product of 1 and any number is that number: 1t = t. If we substitute 40 for t in 2t = 80, we obtain the true statement 80 = 80. This verifies that 40 is the solution. The solution s (b) To isolatet on the right side, we use the division property of equality. We can undo the multiplication by –8.6 by dividing bo -6.02 = -8.6t This is the equation to solve. -6.02 -8.6t
Solve: (a) 2t = 80 (b) -6.02 = -8.6t. We will use a property of equality to isolate the variable on one side of the equation. To solve the original equation, we want to find a simpler equivalent equation of the form t = a number or a number = t, whose solution is obvious. (a) To isolatet on the left side, we use the division property of equality. We can undo the multiplication by 2 by dividing both sic 2t = 80 This is the equation to solve. 2t 80 Use the division property of equality: Divide both sides by 2. %3D 2 it = Simplify 4 by removing the common factor of 2 in the numerator and denominator: = 1. t = 40 The product of 1 and any number is that number: 1t = t. If we substitute 40 for t in 2t = 80, we obtain the true statement 80 = 80. This verifies that 40 is the solution. The solution s (b) To isolatet on the right side, we use the division property of equality. We can undo the multiplication by –8.6 by dividing bo -6.02 = -8.6t This is the equation to solve. -6.02 -8.6t
Algebra and Trigonometry (6th Edition)
6th Edition
ISBN:9780134463216
Author:Robert F. Blitzer
Publisher:Robert F. Blitzer
ChapterP: Prerequisites: Fundamental Concepts Of Algebra
Section: Chapter Questions
Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...
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Equations and Inequations
Equations and inequalities describe the relationship between two mathematical expressions.
Linear Functions
A linear function can just be a constant, or it can be the constant multiplied with the variable like x or y. If the variables are of the form, x2, x1/2 or y2 it is not linear. The exponent over the variables should always be 1.
Question
![Solve:
(a)
2t = 80 (b)
-6.02 = -8.6t.
We will use a property of equality to isolate the variable on one side of the
equation.
To solve the original equation, we want to find a simpler equivalent equation of the
form t = a number or a number =
t, whose solution is obvious.
(a) To isolate t on the left side, we use the division property of equality. We can undo the multiplication by 2 by dividing both sides of the equation by 2.
2t
80
This is the equation to solve.
2t
80
Use the division property of equality: Divide both sides by 2.
2t
2
Simplify
by removing the common factor of 2 in the numerator and denominator:
2
1t =
1.
2
t = 40
The product of 1 and any number is that number: 1t = t.
If we substitute 40 for t in 2t = 80, we obtain the true statement 80 = 80. This verifies that 40 is the solution. The solution set is {40}.
(b) To isolatet on the right side, we use the division property of equality. We can undo the multiplication by -8.6 by dividing both sides by -8.6.
-6.02
= -8.6t
This is the equation to solve.
-6.02
-8.6t
Use the division property of equality: Divide both sides by -8.6.
-8.6
= t
Do the division: 8.6 6.02. The quotient of two negative numbers is positive.
The solution is 0.7. Verify that this is correct by checking.
Solve.
(a)
17x = 204
X =
(b)
9.24 = -0.4r
r =](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F3c69c300-2450-4b46-97af-8596007e11b9%2F4eaee0c2-474e-41c0-bcc5-4852b6985ce9%2F4fzwfn_processed.png&w=3840&q=75)
Transcribed Image Text:Solve:
(a)
2t = 80 (b)
-6.02 = -8.6t.
We will use a property of equality to isolate the variable on one side of the
equation.
To solve the original equation, we want to find a simpler equivalent equation of the
form t = a number or a number =
t, whose solution is obvious.
(a) To isolate t on the left side, we use the division property of equality. We can undo the multiplication by 2 by dividing both sides of the equation by 2.
2t
80
This is the equation to solve.
2t
80
Use the division property of equality: Divide both sides by 2.
2t
2
Simplify
by removing the common factor of 2 in the numerator and denominator:
2
1t =
1.
2
t = 40
The product of 1 and any number is that number: 1t = t.
If we substitute 40 for t in 2t = 80, we obtain the true statement 80 = 80. This verifies that 40 is the solution. The solution set is {40}.
(b) To isolatet on the right side, we use the division property of equality. We can undo the multiplication by -8.6 by dividing both sides by -8.6.
-6.02
= -8.6t
This is the equation to solve.
-6.02
-8.6t
Use the division property of equality: Divide both sides by -8.6.
-8.6
= t
Do the division: 8.6 6.02. The quotient of two negative numbers is positive.
The solution is 0.7. Verify that this is correct by checking.
Solve.
(a)
17x = 204
X =
(b)
9.24 = -0.4r
r =
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