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Chapter 2: Q64. (page 124)

Solve each equation. Check your solution. (Lesson 2 -3)

15=4a−5

Short Answer

Expert verified

The solution isa=5

Step by step solution

01

Step 1. Apply the concept of addition/subtraction and multiplication/division of equality.

We can add or subtract same quantities on both sides of the equation for separating the variable on one side of the equation and the numbers on the other side.

We can multiply or divide by same quantities on both sides of the equation for separating the variable on one side of the equation and the numbers on the other side.

02

Step 2. Solve the equation.

Using concepts mentioned in step -1, we solve equation as shown:

15=4a−54a=15+54a=20a=204a=5

Thus the solution of the given equation isa=5

03

Step 3. Checking the solution.

We substitute the value of the variable into the given equation:

15=4a−515=45−515=20−515=15

Since, the above equation is satisfied, the solution a=5 is verified.

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Most popular questions from this chapter

Translate each sentence into an equation.

Thirty-two divided by y is equal to the product of three and y minus four.

Which equation represents the second step of the solution process?

Step 1: 42x+7−6=3x

Step 2: _________________

Step 3: 5x+28−6=0

Step 4: 5x=−22

Step 5: x=−4.4

  1. 42x−6+7=3x
  2. 42x+1=3x
  3. 8x+7−6=3x
  4. 8x+28−6=3x

Translate the following equation into a sentence:

3g+8=4h−10

Translate the following equation into a sentence:

6d2−7f=8d+f2

Find the value of n. Then name the property used in each step. (Lesson 1-3)

4⋅14=n
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