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Chapter 4: Problem 50

Solve by substitution. Include the units of measurement in the solution. $$ \begin{aligned} &y=\left(\frac{\$ 25}{1 \text { product }}\right) x+\$ 10,000 \\ &y=\left(\frac{\$ 50}{1 \text { product }}\right) x \end{aligned} $$

Short Answer

Expert verified
x = 400 products, y = \$ 20,000.

Step by step solution

01

Set the equations equal

Since both equations equal y, set the right-hand sides of the equations equal to each other: \( \left( \frac{\$ 25}{1 \text{ product}} \right) x + \$ 10,000 = \left( \frac{\$ 50}{1 \text{ product}} \right) x \)
02

Isolate the variable

To isolate x, subtract \( \left( \frac{\$ 25}{1 \text{ product}} \right) x \) from both sides: \( \$ 10,000 = \left( \frac{\$ 50}{1 \text{ product}} \right) x - \left( \frac{\$ 25}{1 \text{ product}} \right) x \)
03

Simplify the equation

Combine like terms: \( \$ 10,000 = \left( \frac{\$ 25}{1 \text{ product}} \right) x \)
04

Solve for x

Divide both sides by \( \frac{\$ 25}{1 \text{ product}} \) to find x: \( x = \frac{\$ 10,000}{\frac{\$ 25}{1 \text{ product}}} \). Thus, \( x = 400 \text{ products} \)
05

Substitute x back into either original equation

Use one of the original equations to find y. For example, \( y = \left( \frac{\$ 25}{1 \text{ product}} \right) x + \$ 10,000 \). Substitute x = 400: \( y = \left( \frac{\$ 25}{1 \text{ product}} \right) (400) + \$ 10,000 = \$ 10,000 + \$ 10,000 = \$ 20,000 \)

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Key Concepts

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

Algebraic Equations
An algebraic equation is a statement that connects two expressions with an equal sign. It's like a balance where both sides need to be equal. For example, in the exercise, we had two equations: \( y=\left(\frac{\$ 25}{1 \text{ product }}\right)x+\$ 10,000 \) and \( y = \left(\frac{\$ 50}{1 \text{ product }}\right)x \). Both of these equations express a relationship between 'y' and 'x'. Algebraic equations can often involve numbers, operations (like addition and multiplication), and variables, which are letters representing unknown values.
Solving Linear Equations
Solving linear equations means finding the value of the variables that make the equation true. To solve a linear equation, you perform operations that help you isolate the variable.
In our exercise, we used substitution to solve the equations. Here's the general process:
  • First, set the equations equal to each other (since both equal 'y').
  • Next, isolate the variable (in this case, 'x').
  • Then, simplify the resulting equation.
  • Finally, solve for the variable by performing arithmetic operations.
Following these steps ensures you correctly find the value of the variable 'x'.
Variable Isolation
Variable isolation is a key step in solving equations. It means getting the variable you are solving for by itself on one side of the equation.
Let's see how we isolated 'x' in the example:
  • We started with \( \left(\frac{}{1\text{product}}\right)x+\$10,000=\left(\frac{\$50}{1\text{product}}\right)x \).
  • We subtracted \( \left(\frac{\$25}{1\text{product}}\right)x \) from both sides, which gave us \( \$10,000 = \left(\frac{\$25}{1\text{product}}\right)x \).
  • Finally, we divided both sides by \(\left(\frac{\$25}{1\text{product}}\right)\) to find 'x.'
This process of steps keeps the equation balanced while isolating our variable.
Units of Measurement in Algebra
Using units of measurement in algebra ensures that your solutions make sense in real-world contexts. In our example, the units were dollars per product and total dollars.
Here鈥檚 why units are useful:
  • They help confirm that the math you鈥檙e doing relates to the real world.
  • They prevent mistakes by ensuring all parts of the equation align correctly.
For instance, when solving \( x = \frac{\$10,000}{\left(\frac{\$25}{1\text{product}}\right)} \), the units checked out, leading us to \( x = 400\text{products}\).
Paying attention to units helps clarify and validate your solutions.

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Most popular questions from this chapter

For exercises \(3-22\), graph. Label the solution region.a $$ \begin{aligned} &x+y \leq 8 \\ &x \geq 2 \end{aligned} $$

A karat describes the percent gold in an alloy (a mixture of metals). $$ \begin{array}{|c|c|} \hline \text { Name of alloy } & \text { Percent gold } \\ \hline \text { 10-karat gold } & 41.7 \% \\ \text { 14-karat gold } & 58.3 \% \\ \text { 18-karat gold } & 75 \% \\ \text { 20-karat gold } & 83.3 \% \\ \text { 24-karat gold } & 100 \% \\ \hline \end{array} $$ Find the amount of 24-karat gold and the amount of silver to mix to make 8 oz of 10 -karat gold. Round to the nearest hundredth.

Pigeon Feed A contains popcorn, whole milo, Canadian peas, whole wheat, maple peas, Austrian peas, and oat groats and is \(14 \%\) protein. Pigeon Feed B contains popcorn, milo, wheat, oat groats, and Red Proso Millet and is \(17 \%\) protein. Find the amount of each feed to mix together to make \(50 \mathrm{lb}\) of a new feed that is \(16.2 \%\) protein. Round to the nearest tenth.

For exercises \(85-88\), the completed problem has one mistake. (a) Describe the mistake in words, or copy down the whole problem and highlight or circle the mistake. (b) Do the problem correctly. Problem: Write a system of equations that represents the relationship of the variables. Mixture A is 6\% juice. Mixture B is \(15 \%\) juice. Mixture B contains \(50 \%\) of the RDA of vitamin C. Find the amount of each mixture needed to make 80 gal of a new drink that is \(11 \%\) juice. Round to the nearest whole number. $$ \begin{aligned} &x=\text { amount of Mixture A } \\ &y=\text { amount of Mixture B } \\ &\text { Incorrect Answer: } x+y=80 \text { gal } \\ &\qquad 0.06 x+0.50 y=(0.11)(80 \mathrm{gal}) \end{aligned} $$

A karat describes the percent gold in an alloy (a mixture of metals). $$ \begin{array}{|c|c|} \hline \text { Name of alloy } & \text { Percent gold } \\ \hline \text { 10-karat gold } & 41.7 \% \\ \text { 14-karat gold } & 58.3 \% \\ \text { 18-karat gold } & 75 \% \\ \text { 20-karat gold } & 83.3 \% \\ \text { 24-karat gold } & 100 \% \\ \hline \end{array} $$ If 9 oz of 14-karat gold jewelry is melted down, find the amount of 20 -karat gold to add to the melted jewelry to create a new alloy that is \(18-k\) arat gold. Find the amount of the new alloy. Round to the nearest tenth.
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