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Chapter 5: Problem 9

Solve each equation. Don't forget to check each of your potential solutions. \(\sqrt{3 y+1}=4\)

Short Answer

Expert verified
The solution is \(y = 5\).

Step by step solution

01

Isolate the Square Root

The equation given is \( \sqrt{3y+1} = 4 \). Since the square root is already isolated on one side of the equation, we can proceed to the next step.
02

Square Both Sides

To eliminate the square root, square both sides of the equation: \((\sqrt{3y+1})^2 = 4^2\). This simplifies to \(3y + 1 = 16\).
03

Solve for y

Now, solve the linear equation. Subtract 1 from both sides: \(3y = 15\). Then, divide both sides by 3: \(y = 5\).
04

Verify the Solution

Substitute \(y = 5\) back into the original equation to check if it is a valid solution. \(\sqrt{3(5) + 1} = \sqrt{15 + 1} = \sqrt{16} = 4\). The left-hand side matches the right-hand side, confirming that \(y = 5\) is a correct solution.

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Key Concepts

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

Linear Equations
Linear equations are mathematical statements that involve variables, coefficients, and constants. They usually take the form of ax + b = c, where "x" is the variable. The goal in solving linear equations is to find the value of the variable that makes the equation true.

Why are they so important?
  • Linear equations are fundamental in algebra and are the building blocks for understanding more complex mathematics.
  • They model real-world situations, helping us to solve problems in science, engineering, and finance.
In our example, after removing the square root in Step 2, we derived a linear equation: 3y + 1 = 16.
We solved this by performing inverse operations, like subtraction and division, to isolate "y".
Square Roots
Square roots are a critical concept in mathematics. When you see a square root symbol (\( \sqrt{} \)), it represents a number which, when multiplied by itself, gives the original number under the root sign.

For example, \( \sqrt{16} = 4 \) because \( 4 \cdot 4 = 16 \). Here's why mastering them is key:
  • They help us in mathematical problem solving, especially in quadratic equations and geometry.
  • Knowing how to manipulate square roots is crucial for simplifying equations.
In the original exercise, we had \( \sqrt{3y + 1} = 4 \). To solve it, we removed the square root by squaring both sides. This operation is pivotal in turning the problem into a simpler linear equation.
Verification of Solutions
Verification of solutions is a crucial step in problem-solving. After obtaining a potential solution, we must check it against the original equation to ensure its correctness.

Why is this verification step crucial?
  • It helps to catch mistakes made during the solving process.
  • It ensures that our solution actually satisfies the original condition posed by the equation.
Let鈥檚 take the value \( y = 5 \) from the original step-by-step solution. By substituting \( y = 5 \) back into \( \sqrt{3y + 1} \), we calculated \( \sqrt{16} = 4 \), which matches the right side of the equation.

This confirms that our solution of \( y = 5 \) is indeed valid, illustrating the importance of always verifying your answer for accuracy.

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Most popular questions from this chapter

Simplify each of the following. Express final results using positive exponents only. For example,\(\left(2 x^{\frac{1}{2}}\right)\left(3 x^{\frac{1}{3}}\right)=6 x^{\frac{5}{6}}\). \(\left(\frac{a^{3}}{b^{-2}}\right)^{-\frac{1}{3}}\)

Simplify each of the following. Express final results using positive exponents only. For example,\(\left(2 x^{\frac{1}{2}}\right)\left(3 x^{\frac{1}{3}}\right)=6 x^{\frac{5}{6}}\). \(\left(\frac{60 a^{\frac{1}{5}}}{15 a^{\frac{3}{4}}}\right)^{2}\)

Use scientific notation and the properties of exponents to help you perform the following operations. \(\sqrt{9,000,000}\)

Perform the indicated operations and express answers in simplest radical form. (See Example 5.) \(\frac{\sqrt[3]{8}}{\sqrt[4]{4}}\)

Use scientific notation and the properties of exponents to help you perform the following operations. \(\frac{(0.0045)(60,000)}{(1800)(0.00015)}\)
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