Problem 64
Question
Write the equation in standard form with integer coefficients. $$y=-\frac{2}{5} x$$
Step-by-Step Solution
Verified Answer
The standard form of the equation with integer coefficients is \(2x+5y=0\).
1Step 1: Write given equation
The given equation is \(y=-\frac{2}{5}x\).
2Step 2: Transformation into standard form
Multiply both sides of the equation by 5 to eliminate the fraction. It gives \(5y=-2x\). Note that typically, the coefficient of x is positive in the standard form.
3Step 3: Arrangement in standard form
Rearrange to get the 'x' term first. This gives us \(2x+5y=0\) which is the standard form of the equation with integer coefficients.
Key Concepts
Integer CoefficientsAlgebraic ExpressionsEquation Transformation
Integer Coefficients
The term 'integer coefficients' refers to the numbers that multiply the variables in an algebraic expression, which are whole numbers and can be positive, negative, or zero. When we represent equations, especially linear equations, in standard form, it’s crucial for the coefficients to be integers. This makes the equations simpler to work with, particularly when solving systems of linear equations or performing operations like addition or subtraction on the equations.
For example, the given exercise emphasizes transforming an equation with a fractional coefficient to one with integer coefficients. Multiplying both sides of the equation by 5 turns \(y = -\frac{2}{5}x\) into \(5y = -2x\), removing fractional parts and resulting in integer coefficients: -2 for 'x' and 5 for 'y'. It's a strategic maneuver that enhances the readability and solvability of the equation.
For example, the given exercise emphasizes transforming an equation with a fractional coefficient to one with integer coefficients. Multiplying both sides of the equation by 5 turns \(y = -\frac{2}{5}x\) into \(5y = -2x\), removing fractional parts and resulting in integer coefficients: -2 for 'x' and 5 for 'y'. It's a strategic maneuver that enhances the readability and solvability of the equation.
Algebraic Expressions
Algebraic expressions are a foundational concept in algebra, comprising numbers, variables, and arithmetic operations. They represent values that can change, which is why we use variables instead of specific numbers. These expressions become powerful tools when they are used to create equations that describe relationships between quantities.
In the context of our exercise, \(y=-\frac{2}{5}x\) is an algebraic expression that describes a linear relationship between 'x' and 'y'. Here, \(y\) is dependent on \(x\), with the fraction \(\frac{2}{5}\) representing the rate of change or slope in this relationship. Understanding how to manipulate these expressions is key to mastering the art of algebra and advancing to more complex mathematical concepts.
In the context of our exercise, \(y=-\frac{2}{5}x\) is an algebraic expression that describes a linear relationship between 'x' and 'y'. Here, \(y\) is dependent on \(x\), with the fraction \(\frac{2}{5}\) representing the rate of change or slope in this relationship. Understanding how to manipulate these expressions is key to mastering the art of algebra and advancing to more complex mathematical concepts.
Equation Transformation
Equation transformation is the process of altering the form of an equation without changing its solution set. This process is extremely valuable as it can transform an equation into a format that is easier to interpret or solve. The standard form for a linear equation is \(Ax + By = C\), where 'A', 'B', and 'C' are integers, and 'A' should be non-negative.
In our example, we took the equation \(y = -\frac{2}{5}x\) and performed a transformation to place it in standard form. We multiplied both sides to eliminate fractions and then rearranged the terms to get \(2x + 5y = 0\) with the 'x' term first, as per convention. This standard form is particularly useful for quickly graphing lines and solving systems of equations.
In our example, we took the equation \(y = -\frac{2}{5}x\) and performed a transformation to place it in standard form. We multiplied both sides to eliminate fractions and then rearranged the terms to get \(2x + 5y = 0\) with the 'x' term first, as per convention. This standard form is particularly useful for quickly graphing lines and solving systems of equations.
Other exercises in this chapter
Problem 64
Solve the equation. Round the result to the nearest hundredth. $$8.75 x+2.16=18.28 x-6.59$$
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Write the number in scientific notation. At the end of 1999 the population of the world was estimated at 6,035,000,000.
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The power generated by a windmill can be modeled by \(w=0.015 s^{3},\) where \(w\) is the power measured in watts and s is the wind speed in miles per hour. Fin
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Simplify the expression \(\frac{x^{-9}}{x^{-3}}\) A. \(\frac{1}{x^{-6}}\) B. \(\frac{1}{x^{6}}\) C. \(x^{-3}\) D. \(x^{3}\)
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