/*! 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} Problem 11 Show how you can solve these equ... [FREE SOLUTION] | 91Ó°ÊÓ

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Show how you can solve these equations by using an undoing process. Check your results by substituting the solutions into the original equations. a. \(-15=-52+1.6 x\) b. \(7-3 x=52\)

Short Answer

Expert verified
a. \(x = 23.125\); b. \(x = -15\). Verified solutions match the original equations.

Step by step solution

01

- Isolate the variable (a)

In the equation \(-15 = -52 + 1.6x\), the goal is to solve for \(x\). Start by adding 52 to both sides to eliminate the constant on the right side. This gives you \(-15 + 52 = 1.6x\) or \(37 = 1.6x\).
02

- Solve for "x" (a)

Divide both sides of the equation \(37 = 1.6x\) by 1.6 to isolate \(x\). The equation becomes \(x = \frac{37}{1.6}\). Calculate this to get \(x = 23.125\).
03

- Verify the solution (a)

Substitute \(x = 23.125\) back into the original equation: \(-15 = -52 + 1.6(23.125)\). Calculate \(1.6 \times 23.125\), which equals 37. Add \(-52\) and 37 to check: \(-15 = -15\). The solution is verified.
04

- Isolate the variable (b)

In the equation \(7 - 3x = 52\), subtract 7 from both sides to eliminate the constant: \(-3x = 52 - 7\), which simplifies to \(-3x = 45\).
05

- Solve for "x" (b)

Divide both sides by -3 to isolate \(x\): \(x = \frac{45}{-3}\). This simplifies to \(x = -15\).
06

- Verify the solution (b)

Substitute \(x = -15\) back into the original equation: \(7 - 3(-15) = 52\). This becomes \(7 + 45 = 52\), which simplifies to \(52 = 52\). The solution is verified.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Isolation of Variables
When solving linear equations, one of the first and most crucial steps is the isolation of variables. This involves rearranging the equation so that the variable we want to solve for stands alone on one side of the equation. Let's break this down with an example.

Consider the equation \( -15 = -52 + 1.6x \). Here, we need to isolate the variable \( x \). To start, we aim to get rid of the constant term, \( -52 \), on the side with the variable. We do this by performing the opposite arithmetic operation. Adding 52 to both sides changes the equation to \( -15 + 52 = 1.6x \). This simplifies to \( 37 = 1.6x \).

In the equation \( 7 - 3x = 52 \), we will isolate \( x \) by first subtracting 7 from both sides of the equation. This simplifies to \( -3x = 45 \).

In these examples, removing the constant terms is key. It sets the stage for you to deal directly with the variable you're solving for.
Verification of Solutions
Once you've found potential solutions by isolating and solving for the variable, it's essential to verify these solutions. Verification involves substituting the value back into the original equation to ensure it holds true.

For instance, after isolating and solving for \( x \) in our first example, we found \( x = 23.125 \). To verify, substitute \( x \) back into the original equation, \( -15 = -52 + 1.6x \). After calculating \( 1.6 \times 23.125 = 37 \) and substituting, you'll find that \( -52 + 37 = -15 \). Since both sides of the equation are equal, the solution is verified.

Similarly, for the equation \( 7 - 3x = 52 \), substituting \( x = -15 \) yields \( 7 - 3(-15) = 52 \). This becomes \( 7 + 45 = 52 \), confirming that \( 52 = 52 \). Both verifications show that the solutions are correct.

Verification is a critical step, as it ensures accuracy and confidence in your computed solutions.
Arithmetic Operations
Arithmetic operations are fundamental in manipulating equations. The key arithmetic operations include addition, subtraction, multiplication, and division. Each plays a critical role in solving linear equations.

During equation solving, we often use arithmetic operations to eliminate constants or coefficients attached to variables. This is evident in how we isolated variables in earlier examples. For the equation \( 37 = 1.6x \), division is used to isolate \( x \). Divide both sides by 1.6, resulting in \( x = \frac{37}{1.6} \).

In another scenario, subtracting 7 from both sides in the equation \( 7 - 3x = 52 \) was necessary to bring the constant to the other side, simplifying it to \( -3x = 45 \). Then, we use division again to solve for \( x \).

By mastering these basic arithmetic operations, you can efficiently manipulate and solve equations, making them a vital part of any math toolkit.

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