Problem 11
Question
In Exercises \(1-21,\) solve the equation for the variable. $$ \sqrt{3 x-2}+1=10 $$
Step-by-Step Solution
Verified Answer
Question: Solve the equation \(\sqrt{3x - 2} + 1 = 10\) for x.
Answer: \(x = \frac{83}{3}\)
1Step 1: Isolate the square root term
Subtract 1 from both sides of the equation \(\sqrt{3x - 2} + 1 = 10\).
$$
\sqrt{3x - 2} = 9
$$
2Step 2: Eliminate the square root
Square both sides of the equation \(\sqrt{3x - 2} = 9\) to eliminate the square root.
$$
(3x - 2) = (9)^2
$$
3Step 3: Simplify the equation
Calculate the square \(9^2\) and expand the left side of the equation.
$$
3x - 2 = 81
$$
4Step 4: Solve for x
Add 2 to both sides of the equation \(3x - 2 = 81\).
$$
3x = 83
$$
Now, divide both sides of the equation by 3 to find x.
$$
x = \frac{83}{3}
$$
So, the solution to the equation is \(x = \frac{83}{3}\).
Key Concepts
Square RootsIsolation of VariablesAlgebraic Manipulation
Square Roots
A square root is a value that, when multiplied by itself, gives the original number. For example, the square root of 9 is 3, because 3 multiplied by 3 equals 9. In terms of equations, finding the square root of a number essentially reverses the process of squaring that number. In our exercise, we started with an equation involving a square root: \( \sqrt{3x - 2} + 1 = 10 \). To solve for \( x \), it was beneficial to first isolate the square root term, \( \sqrt{3x - 2} \), before attempting to eliminate the square root from the equation. Once isolated, squaring both sides allowed us to proceed without the square root, transforming it into a standard algebraic expression.
Isolation of Variables
Isolation of variables is a fundamental step in solving equations. It involves rearranging an equation so that one variable stands alone on one side of the equation. In the given exercise, our goal was to solve for the variable \( x \). The equation \( \sqrt{3x - 2} + 1 = 10 \) was initially structured with additional terms that needed to be removed. By subtracting 1 from both sides, we isolated the square root term: \( \sqrt{3x - 2} = 9 \). This isolation allows us to directly address the term containing our variable of interest, simplifying the problem and making further steps more straightforward.
Algebraic Manipulation
Algebraic manipulation is all about rearranging and simplifying expressions to solve for unknowns. After isolating the square root term in our equation, the next step required algebraic manipulation to get rid of the square root. We squared both sides of \( \sqrt{3x - 2} = 9 \) to eliminate the square root, resulting in \( 3x - 2 = 81 \). Continuing with algebraic techniques, we simplified the equation by first adding 2 to both sides, which gave \( 3x = 83 \). Finally, by dividing each side by 3, we solved for \( x \), resulting in \( x = \frac{83}{3} \). These actions involve routine algebraic operations that are key to solving virtually any equation, regardless of complexity.
Other exercises in this chapter
Problem 10
Write the expression as a constant times a power of a variable. Identify the coefficient and the exponent. $$ \frac{\sqrt{x}}{\sqrt[3]{x}} $$
View solution Problem 11
In Exercises \(9-12,\) write a formula representing the function. The radius, \(r,\) of a circle is proportional to the square root of the area, \(A\).
View solution Problem 11
Write the expression as a constant times a power of a variable. Identify the coefficient and the exponent. $$ \frac{x^{4}}{4 \sqrt{x^{2}}}, x>0 $$
View solution Problem 12
The radius, \(r,\) of a circle is proportional to the square root of the area, \(A\). Kinetic energy, \(K\), is proportional to the square of velocity, \(v .\)
View solution