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Chapter 2: Problem 112

Why can't you multiply just one side of an equation by something when algebraically rearranging the equation?

Short Answer

Expert verified
When algebraically rearranging an equation, it is crucial to maintain the balance of the equation. If you multiply just one side of the equation, it breaks the balance and results in an incorrect solution. In our example, multiplying only the left side of the equation \(x+2=5\) by 2 led to a different and incorrect value for x. To keep the equation balanced, always perform the same operation on both sides of the equation.

Step by step solution

01

Example Equation

Consider the equation \(x + 2 = 5\).
02

Multiply Just One Side

Suppose we decide to multiply only the left side of the equation by 2. Then, our equation would become \(2(x + 2) = 5\).
03

Simplify the Equation

Now, let's simplify the equation we created by multiplying only the left side by 2. \(2(x + 2) = 2x + 4 = 5\)
04

Attempt to Solve the Equation

Now, let's attempt to solve the equation for x: \(2x + 4 = 5\) Subtract 4 from both sides: \(2x = 1\) Divide by 2: \(x = 0.5\)
05

Compare the Solutions

With our modified equation, we've determined x = 0.5. However, if we solve the original equation: \(x + 2 = 5\) Subtract 2 from both sides: \(x = 3\)
06

Conclusion

In this example, you can see that when we multiplied just one side of the equation, it resulted in a different (and incorrect) value for x. Maintaining the balance of the equation is crucial for finding the correct algebraic solutions, which is why you must perform the same operation on both sides of an equation.

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Most popular questions from this chapter

You measure one edge of a cube using a meterstick marked in centimeters. Unfortunately, the edge is longer than \(1 \mathrm{~m}\). You mark the \(1-\mathrm{m}\) point on the cube edge with a pen and then, using a \(15-\mathrm{cm}\) ruler marked in millimeters, measure the remaining distance to be \(1.40 \mathrm{~cm}\). (a) What is the length of the edge in centimeters? (b) What is the volume of the cube in cubic centimeters? (Remember, the lengths of all edges of a cube are equal.) Watch your significant figures. Use scientific notation if you have to. (c) The cube has a mass of \(111 \mathrm{~kg} .\) What is its density in grams per milliliter? Watch your significant figures.

Use a scientific calculator to do the following calculations. Express each answer in scientific notation and to the correct number of significant figures. (a) \(9.865 \times 10^{3}+8.61 \times 10^{2}\) (b) \(\frac{\left(6.626 \times 10^{23}\right) \times\left(3.00 \times 10^{8}\right)}{4.5 \times 10^{-7}}\) (c) \(\frac{5.6200 \times 10^{-9}}{3.821 \times 10^{9}}\) (d) \(\frac{4.5600 \times 10^{3}-2.91 \times 10^{1}}{5}\), where the 5 is an exact number

The density of water at \(4.00^{\circ} \mathrm{C}\) is \(1.00 \mathrm{~g} / \mathrm{mL}\). The density of ice at \(0{ }^{\circ} \mathrm{C}\) is \(0.917 \mathrm{~g} / \mathrm{mL}\). Water is different from most other substances in that the solid phase (ice) is less dense than the liquid phase. Explain why this characteristic makes ice fishing possible.

Convert: (a) \(23.0{ }^{\circ} \mathrm{C}\) to \(\mathrm{K}\) (b) \(98.6^{\circ} \mathrm{F}\) to \({ }^{\circ} \mathrm{C}\) (c) \(296 \mathrm{~K}\) to \({ }^{\circ} \mathrm{F}\) (d) \(32^{\circ} \mathrm{F}\) to \(\mathrm{K}\) (e) \(523 \mathrm{~K}\) to \({ }^{\circ} \mathrm{C}\) (f) \(38^{\circ} \mathrm{C}\) to \({ }^{\circ} \mathrm{F}\)

Which contains more matter: a \(50-\mathrm{g}\) block of lead or a \(50-\mathrm{g}\) block of gold? Explain.
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