Question

Solve each equation for the variable: (a) \(x=\frac{212-32}{1.8}\) (b) \(8.9 \frac{\mathrm{g}}{\mathrm{mL}}=\frac{40.90 \mathrm{~g}}{x}\) (c) \(72^{\circ} \mathrm{F}=1.8 x+32\)

Short answer

a) \(x = 100\)b) \(x \approx 4.6\)c) \(x \approx 22.2\)

Step-by-step solution

Solve Equation (a)

Given equation is: \( x = \frac{212 - 32}{1.8} \)First, subtract 32 from 212: \( 212 - 32 = 180 \) So the equation becomes: \( x = \frac{180}{1.8} \)Now, divide 180 by 1.8: \( x = 100 \)

Solve Equation (b)

Given equation is: \( 8.9 \frac{\mathrm{g}}{\mathrm{mL}} = \frac{40.90 \mathrm{~g}}{x} \)First, rewrite the equation to solve for \(x\): \( x = \frac{40.90 \mathrm{~g}}{8.9 \frac{\mathrm{g}}{\mathrm{mL}}} \)Now, divide 40.90 by 8.9: \( x \approx 4.6 \mathrm{~mL} \)

Solve Equation (c)

Given equation is: \( 72^{\circ} \mathrm{F} = 1.8 x + 32 \)First, subtract 32 from both sides: \( 72 - 32 = 1.8 x \)So the equation becomes: \( 40 = 1.8 x \)Now, divide both sides by 1.8: \( x \approx 22.2 \)

Key concepts

Variable Isolation

In the realm of solving equations, one of the fundamental practices is variable isolation. This means manipulating the equation in such a way that the variable of interest stands alone on one side of the equation.
In our examples, we observed this in each of the given steps. For instance, in equation (c), we needed to isolate the variable \(x\) from the equation 72°F = 1.8x + 32.
Here's how it works:

• First, subtract 32 from both sides to move the constant term away from the variable: 72 - 32 = 1.8x.
• This simplified the equation to 40 = 1.8x which is easier to solve.
• Next, to completely isolate \(x\), divide both sides by 1.8, resulting in \(x \rightarrow 22.2\).

The aim is always the same - use arithmetic operations to ensure \(x\) (or any other variable) is by itself on one side of the equation.

Unit Conversion

Unit conversion is another key concept in these examples. It means converting different units of measure to help solve equations more easily and accurately.
Let's take a look at equation (b): 8.9 \(\frac{g}{mL}\) = \(\frac{40.90 \text{ g}}{x}\):

• Here, g (grams) and mL (milliliters) are units of measure that need to be consistent for the calculations to make sense.
• By rewriting the equation as x = \(\frac{40.90 \text{ g}}{8.9 \frac{g}{mL}}\), we ensure that g cancels out, leaving mL, which is what we need to find.
• This results in: \(x \rightarrow 4.6\) mL.

Without proper unit conversions, solving equations that deal with measurements can lead to confusing or incorrect results.

Linear Equations

Linear equations are equations where the highest power of the variable is one. In simpler terms, they form straight lines when graphed.
For instance, in equation (a), \(x = \frac{212 - 32}{1.8}\), we simplified it step by step:

• First we performed the subtraction in the numerator to get x = \(\frac{180}{1.8}\).
• Then we divided 180 by 1.8, leading to \(x = 100\).

Each of these equations can be rewritten in the form of \(ax + b = c\) where \(a\), \(b\), and \(c\) are constants and \(x\) is the variable.
Understanding the concept of linear equations is crucial as it forms the basis for more complex types of equations and mathematical modeling.