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Chapter 4: Problem 20

Solve the equation algebraically. Check your solution graphically. $$5 x+3=-2$$

Short Answer

Expert verified
The solution to the equation \(5x + 3 = -2\) is \(x = -1\), checked graphically by confirming the line \(y = 5x+3\) intersects the x-axis at x=-1.

Step by step solution

01

Isolating 'x' Algebraically

We start with the equation \(5x + 3 = -2\). The goal is to isolate 'x'. To do so, first, we subtract 3 from both sides of the equation to remove the +3 on the left. This gets us: \(5x = -2-3\), which simplifies to \(5x = -5\).
02

Solving for 'x'

Next, we need to solve for 'x'. The current term with 'x' (\(5x\)) indicates 5 times 'x', so we need to undo that by doing the opposite - in this case, division. We divide both sides of the equation by 5. This gives us: \(x = -5/5\), which further simplifies to \(x = -1\).
03

Checking the solution graphically

To check the solution graphically, create a graph with the equation \(y = 5x + 3\). The solution to the initial equation is the x-coordinate at which the line crosses the x-axis. If we did Q1 and Q2 correctly, the line should intersect the x-axis at x=-1.

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Key Concepts

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

Algebraic Methods for Solving Linear Equations
When students encounter a linear equation such as 5x + 3 = -2, algebraic methods provide a structured approach to finding the solution. The key principle in algebra is to perform the same operation on both sides of the equation to keep it balanced.

Let's break down how this works:
  • Identify the variable you're solving for 鈥� in this case, it's 'x'.
  • Apply the inverse operation to isolate the term with the variable. If the variable is being added to or subtracted from a number, do the opposite (subtract or add) that number from both sides.
  • If the variable is multiplied by a number, divide both sides by that number; if divided, multiply.
Once 'x' is isolated and you have a simpler equation like x = -1, you have found your solution. It鈥檚 critical to understand each algebraic step because it teaches how to maintain balance in equations, a fundamental concept in algebra.
Graphical Solutions to Linear Equations
Graphical solutions are a visual representation of algebraic equations and can serve as a great tool for checking the work you've done using algebraic methods. To graph the equation 5x + 3 = -2, or in graphing format, y = 5x + 3, follow these steps:
  • Start by plotting the y-intercept, which is the constant term in the equation (in this case, +3) on the y-axis.
  • Use the slope (the coefficient of x, which is 5) to determine the rise over run, helping you plot another point.
  • Draw a line through these points, which represents the equation.
The graph will help you visualize where the equation crosses the x-axis (x-intercept), which reflects the solution of the equation. In an accurately drawn graph, the x-intercept should match the solution found using algebraic methods.
Isolation of Variables in Linear Equations
Isolation of the variable is one of the fundamental techniques in solving linear equations. The main goal is to have the variable you're looking for by itself on one side of the equal sign. To isolate the variable 鈥檟鈥� in an equation like 5x + 3 = -2:
  1. Subtract or add constants to both sides to remove them from the side with the variable.
  2. Once you have only terms with the variable on one side, use multiplication or division to get the variable by itself.
This technique not only applies to simple linear equations but also forms the foundation for solving more complex algebraic equations. Understanding variable isolation is crucial in algebra and helps students in recognizing how equations are balanced and manipulated to arrive at a solution.

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Most popular questions from this chapter

The table shows information about the top 7 winners of the 1997 Masters Tournament in Augusta, Georgia. Graph the relation. Is the money earned a function of the score? Explain. $$ \begin{array}{|l|c|c|c|c|c|c|c|}\hline \text { Score } & 270 & 282 & 283 & 284 & 285 & 285 & 286 \\\\\hline \text { Prize (s) } & 486,000 & 291,600 & 183,600 & 129,600 & 102,600 & 102,600 & 78,570 \\\\\hline\end{array} $$

Apply the distributive property. $$-2(d-5)$$

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Evaluate the expression. (Review 2.1 ) $$\left|\frac{-2}{3}\right|$$
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