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Chapter 1: Problem 16

If \(\frac{1}{4} x=5-\frac{1}{2} y,\) what is the value of \(x+2 y ?\)

Short Answer

Expert verified
The value of x + 2y is 20.

Step by step solution

01

- Isolate x

Given the equation \(\frac{1}{4}x = 5 - \frac{1}{2}y\), multiply both sides by 4 to isolate x: \(\frac{1}{4}x \times 4 = (5 - \frac{1}{2}y) \times 4\). This simplifies to \(x = 20 - 2y\).
02

- Express x in terms of y

From Step 1, we have \(x = 20 - 2y\).
03

- Find the value of x + 2y

To find the value of \(x + 2y\), substitute \(x\) from the expression obtained in Step 2: \(x + 2y = (20 - 2y) + 2y\).
04

- Simplify the expression

Simplify the expression \(20 - 2y + 2y\). Since \(-2y + 2y = 0\), we have \(20\).

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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 involves variables and constants combined using arithmetic operations. In this exercise, we start with the equation \(\frac{1}{4}x = 5 - \frac{1}{2}y\). The goal is to solve for the unknowns, which are the variables x and y. Algebraic equations can be linear, quadratic, or higher-order, but this one is linear because it involves x and y to the first power.
variable isolation
Variable isolation means rearranging an equation to get one variable alone on one side. In Step 1, to isolate x, we multiply both sides by 4.
This cancels out the fraction on the left, making it easier to solve for x. Always perform the same operation on both sides to keep the equation balanced.
This gives us: \[ \frac{1}{4} x \times 4 = (5 - \frac{1}{2} y) \times 4 \]
which simplifies to: \[ x = 20 - 2y \]
Now, x is isolated.
substitution method
The substitution method involves replacing one variable with an expression involving another variable. After isolating x, we have:
x = 20 - 2y
We substitute this back into our equation to find the combined value of x and 2y. The substitution step is crucial for connecting different parts of the problem and simplifying it.
simplification
Simplification means making an equation as basic as possible. From the expression: \[ x + 2y = (20 - 2y) + 2y \]
we combine like terms. The -2y and 2y terms cancel out, leaving us with:
\[ x + 2y = 20 \]
This final simplification step can make equations much easier to understand and solve.
It's always good to double-check your work for accuracy.

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Most popular questions from this chapter

If $$\frac{2}{5}(5 x)+2(x-1)=4(x+1)-2,$$ what is the value of \(x ?\) $$ \begin{array}{l}{\text { (A) } x=-2} \\ {\text { (B) } x=2} \\ {\text { (C) There is no value of } x \text { for which the equation is }} \\ {\text { true. }} \\ {\text { (D) There are infinitely many values of } x \text { for which }} \\ {\text { the equation is true. }}\end{array} $$

Which of the following are solutions to the quadratic equation \((x-1)^{2}=\frac{4}{9} ?\) $$ \begin{array}{l}{\text { (A) } x=-\frac{5}{3}, x=\frac{5}{3}} \\ {\text { (B) } x=\frac{1}{3}, x=\frac{5}{3}} \\ {\text { (C) } x=\frac{5}{9}, x=\frac{13}{9}} \\ {\text { (D) } x=1 \pm \sqrt{\frac{2}{3}}}\end{array} $$

$$x+3 y \leq 18$$ $$2 x-3 y \leq 9$$ If \((a, b)\) is a point in the solution region for the system of inequalities shown above and \(a=6,\) what is the minimum possible value for \(b\) ?

The percent increase from 5 to 12 is equal to th percent increase from 12 to what number? $$ \begin{array}{l}{\text { (A) } 16.8} \\ {\text { (B) } 19.0} \\ {\text { (C) } 26.6} \\ {\text { (D) } 28.8}\end{array} $$

A voltage divider is a simple circuit that converts a large voltage into a smaller one. The figure above shows a voltage divider that consists of two resistors that together have a total resistance of 294 ohms. To produce the desired voltage of 330 volts, \(R_{2}\) must be 6 ohms less than twice \(R_{1}\) . Solving which of the following systems of equations gives the individual resistances for \(R_{1}\) and \(R_{2} ?\) $$ (A)\left\\{\begin{array}{l}{R_{2}=2 R_{1}-6} \\\ {R_{1}+R_{2}=294}\end{array}\right. $$ $$ (B)\left\\{\begin{array}{l}{R_{1}=2 R_{2}+6} \\\ {R_{1}+R_{2}=294}\end{array}\right. $$ $$ (C)\left\\{\begin{array}{l}{R_{2}=2 R_{1}-6} \\\ {R_{1}+R_{2}=\frac{294}{330}}\end{array}\right. $$ $$ (D)\left\\{\begin{array}{l}{R_{1}=2 R_{2}+6} \\\ {R_{1}+R_{2}=330}(294)\end{array}\right. $$
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