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Chapter 7: Problem 11

In Exercises \(1-21,\) solve the equation for the variable. $$ \sqrt{3 x-2}+1=10 $$

Short Answer

Expert verified
Question: Solve the equation \(\sqrt{3x - 2} + 1 = 10\) for x. Answer: \(x = \frac{83}{3}\)

Step by step solution

01

Isolate the square root term

Subtract 1 from both sides of the equation \(\sqrt{3x - 2} + 1 = 10\). $$ \sqrt{3x - 2} = 9 $$
02

Eliminate the square root

Square both sides of the equation \(\sqrt{3x - 2} = 9\) to eliminate the square root. $$ (3x - 2) = (9)^2 $$
03

Simplify the equation

Calculate the square \(9^2\) and expand the left side of the equation. $$ 3x - 2 = 81 $$
04

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 \(x = \frac{83}{3}\).

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Key Concepts

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

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.

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Most popular questions from this chapter

Without solving them, say whether the equations in Exercises \(27-42\) have (i) One positive solution (ii) One negative solution (iii) One solution at \(x=0\) (iv) Two solutions (v) \(\quad\) Three solutions (vi) No solution Give a reason for your answer. $$ x^{-3}=-8 $$

Without solving them, say whether the equations in Exercises \(27-42\) have (i) One positive solution (ii) One negative solution (iii) One solution at \(x=0\) (iv) Two solutions (v) \(\quad\) Three solutions (vi) No solution Give a reason for your answer. $$ x^{-2}=4 $$

Suppose \(c\) is inversely proportional to the square of \(d\). If \(c=50\) when \(d=5\), find the constant of proportion-ality and write the formula for \(c\) in terms of \(d\). Use your formula to find \(c\) when \(d=7\).

In Exercises \(43-48\), what operation transforms the first equation into the second? Identify any extraneous solutions and any solutions that are lost in the transformation. $$ \begin{aligned} r^{2}+3 r &=7 r \\ r+3 &=7 \end{aligned} $$

Without solving them, say whether the equations in Problems \(43-56\) have a positive solution \(x=a\) such that (i) \(a>1\) (ii) \(\quad a=1\) (iii) \(\quad 0
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