Problem 43
Question
In Exercises \(43-50,\) solve each equation for \(x .\) $$y=(a+b) x$$
Step-by-Step Solution
Verified Answer
The solution to the equation \(y=(a+b) x\) for \(x\) is \(x = \frac{y}{a + b}\), assuming \(a + b\) is not equal to zero.
1Step 1: Identify the equation
The equation given is \(y = (a + b) x\), where we are asked to solve for \(x\).
2Step 2: Isolate x
To isolate \(x\), divide both sides of the equation by the quantity \(a + b\), ensuring that it is not equal to zero. So, the equation becomes \(x = \frac{y}{a + b}\).
Key Concepts
Isolate VariableLinear EquationsAlgebraic Manipulation
Isolate Variable
Isolating the variable is a fundamental process in solving algebraic equations. It involves rearranging the equation so that the variable of interest, often represented as 'x', stands alone on one side of the equation, with all other terms on the opposite side. This is typically achieved through a combination of operations such as addition, subtraction, multiplication, division, and the use of distributive properties.
For instance, in the given equation from the exercise, \(y = (a + b) x\), the goal is to isolate \(x\). Since \(x\) is being multiplied by the quantity \(a + b\), you perform the opposite operation, which is division, to both sides of the equation in order to get \(x\) by itself. The crucial aspect to remember is that whatever operation is applied to one side of the equation must also be applied to the other, ensuring the equation remains balanced. By divinding both sides by \(a + b\), we isolated the variable \(x\) successfully and obtained \(x = \frac{y}{a + b}\).
For instance, in the given equation from the exercise, \(y = (a + b) x\), the goal is to isolate \(x\). Since \(x\) is being multiplied by the quantity \(a + b\), you perform the opposite operation, which is division, to both sides of the equation in order to get \(x\) by itself. The crucial aspect to remember is that whatever operation is applied to one side of the equation must also be applied to the other, ensuring the equation remains balanced. By divinding both sides by \(a + b\), we isolated the variable \(x\) successfully and obtained \(x = \frac{y}{a + b}\).
Linear Equations
Linear equations form the backbone of algebra and are the simplest type of equations to solve. A linear equation is one in which each term is either a constant or the product of a constant and the first power of a variable. These equations graph as straight lines, hence the name 'linear.'
The standard form of a linear equation is \(Ax + By = C\), where \(A\), \(B\), and \(C\) are constants. In the exercise, the equation given is linear since it can be rearranged into this form with \(y\) on one side and terms involving \(x\) on the other. Linear equations are solved by isolating the variable of interest, and the solution represents all the points that lie on the line corresponding to the equation in a two-dimensional space.
The standard form of a linear equation is \(Ax + By = C\), where \(A\), \(B\), and \(C\) are constants. In the exercise, the equation given is linear since it can be rearranged into this form with \(y\) on one side and terms involving \(x\) on the other. Linear equations are solved by isolating the variable of interest, and the solution represents all the points that lie on the line corresponding to the equation in a two-dimensional space.
Algebraic Manipulation
Algebraic manipulation is the art of reshaping equations using algebraic rules to make them easier to understand or solve. Essential skills in algebraic manipulation include combining like terms, using the distributive property, factoring, expanding expressions, and working with fractions.
In the context of the exercise, manipulating the equation \(y = (a + b) x\) calls for a simple division to both sides to isolate \(x\). It is a direct example of algebraic manipulation, demonstrating how an equation can be transformed in a way that uncovers the value of a variable. As students practice and become more familiar with these techniques, they will find that what once seemed like complex puzzles will become manageable, orderly sets of operations easily applied to a wide array of problems.
In the context of the exercise, manipulating the equation \(y = (a + b) x\) calls for a simple division to both sides to isolate \(x\). It is a direct example of algebraic manipulation, demonstrating how an equation can be transformed in a way that uncovers the value of a variable. As students practice and become more familiar with these techniques, they will find that what once seemed like complex puzzles will become manageable, orderly sets of operations easily applied to a wide array of problems.
Other exercises in this chapter
Problem 43
Solve each equation using both the addition and multiplication properties of equality. Check proposed solutions. $$3 z=-2 z-15$$
View solution Problem 43
Solve each equation. Using the addition property of equality. Be sure to check your proposed solutions. $$-3.7+m=-3.7$$
View solution Problem 43
Solve equation and check your proposed solution. Begin your work by rewriting each equation without fractions. \(\frac{3 x}{4}-3=\frac{x}{2}+2\)
View solution Problem 44
Including \(8 \%\) sales tax, a bed-and-breakfast inn charges 172.80 dollar per night. Find the inn's nightly cost before the tax is added.
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