/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Q. 13 In the following exercise, solve... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 2: Q. 13 (page 215)

In the following exercise, solve each equation using the Subtraction and Addition Properties of Equality.
a鈭�45=76

Short Answer

Expert verified

a=121

Step by step solution

01

Step 1. Given information

In the following exercise, solve each equation using the Subtraction and Addition Properties of Equality.
a鈭�45=76

02

Step 2. Explanation

鈬�a鈭�45=76Add45eachsides鈬�a鈭�45+45=76+45Simplify鈬�a=76+45鈬�a=121

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Solve: 鈭�4(q鈭�2)鈭�8=24.

In the following exercise, solve each equation using the Subtraction and Addition Properties of Equality.
x+0.93=鈭�4.

Solve56=鈭�83r.

Solve: 18鈭�22=12x鈭�x鈭�4x.

Translate and solve: n divided by 7 is equal to 鈭�21 .
See all solutions

Recommended explanations on Math Textbooks

Geometry

Read Explanation

Applied Mathematics

Read Explanation

Pure Maths

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Logic and Functions

Read Explanation

Statistics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.