Problem 26
Question
Solve the equation and check your answer. $$ -3(5-x)-(x-2)=7 x-2 $$
Step-by-Step Solution
Verified Answer
The solution is \( x = \frac{-11}{5} \).
1Step 1: Distribute and Simplify
Begin by distributing the terms on the left side of the equation: \[ -3(5-x) -(x-2) = -15 + 3x - x + 2 \]Combine like terms:\[ 2x - 13 \] So the equation becomes:\[ 2x - 13 = 7x - 2 \]
2Step 2: Move Variables to One Side
Subtract \(2x\) from both sides to begin isolating the variable \(x\):\[ -13 = 7x - 2x - 2 \]This simplifies to:\[ -13 = 5x - 2 \]
3Step 3: Isolate the Variable Term
Add 2 to both sides to isolate the \(5x\) term:\[ -13 + 2 = 5x \]This simplifies to:\[ -11 = 5x \]
4Step 4: Solve for x
Divide each side by 5 to solve for \(x\):\[ x = \frac{-11}{5} \]
5Step 5: Check the Solution
Substitute \(x = \frac{-11}{5}\) back into the original equation to verify the solution:- Compute the left side: \[ -3\left(5 - \frac{-11}{5}\right) - \left(\frac{-11}{5} - 2\right) \] Simplify each part: \[ -3\left(\frac{25 + 11}{5}\right) - \left(\frac{-11 - 10}{5}\right) \] \[ -3\cdot\frac{36}{5} + \frac{21}{5} \] \[ -\frac{108}{5} + \frac{21}{5} = \frac{-87}{5} \]- Compute the right side: \[ 7 \cdot \frac{-11}{5} - 2 \] \[ \frac{-77}{5} - \frac{10}{5} = \frac{-87}{5} \]Both sides equal, confirming the solution is correct.
Key Concepts
Distributive PropertyCombining Like TermsIsolation of VariablesChecking Solutions
Distributive Property
When solving equations, the distributive property is a valuable tool. It's used to eliminate parentheses by multiplying a single term outside the parenthesis by each term inside it. This allows us to simplify expressions and make them easier to work with. For example, in the equation \( -3(5-x)-(x-2) = 7x-2 \), we apply the distributive property by multiplying \(-3\) by both \(5\) and \(-x\), which gives us \(-15\) and \(3x\).
- The minus sign in front of \((x-2)\) impacts both terms, turning it into \(-x+2\).
- This step is crucial for breaking down complex expressions into simpler parts.
Combining Like Terms
Once we've used the distributive property to expand the terms, the next step in our process is combining like terms. "Like terms" are terms in an expression that have identical variable parts, although they may have different coefficients. In our example, \(-15 + 3x - x + 2\), we can combine \(3x - x\) as these are both terms involving the variable \(x\). This simplifies to \(2x\).
- Similarly, the constant terms \(-15\) and \(+2\) can be combined, becoming \(-13\).
- The new expression becomes \(2x - 13\).
Isolation of Variables
The isolation of variables is a pivotal step in solving equations, especially when solving for an unknown variable like \(x\). The goal is to get \(x\) by itself on one side of the equation. Starting from \(2x - 13 = 7x - 2\), we move terms around to isolate \(x\). Subtract \(2x\) from both sides, resulting in \(-13 = 5x - 2\). Following this, we add \(2\) to both sides to clear out the constant term next to the variable term, giving \(-11 = 5x\).
- Through these manipulations, we progressively center \(x\) on one side.
- The final step is to solve for \(x\) by dividing both sides by \(5\), leading to \(x = \frac{-11}{5}\).
Checking Solutions
After finding a solution to an equation, checking that solution is vital to ensure its correctness. To do this, substitute the found value back into the original equation and see if both sides remain equal. Begin by substituting \(x = \frac{-11}{5}\) into \(-3(5-x)-(x-2)=7x-2\). Calculate each side separately.
- The left side simplifies to \(\frac{-87}{5}\) after computing \(-3\left(\frac{36}{5}\right)+\frac{21}{5}\).
- The right side also equals \(\frac{-87}{5}\) upon evaluating \(7 \cdot \frac{-11}{5} - 2\).
Other exercises in this chapter
Problem 25
Exercises \(19-32:\) Graph the linear function by hand. Identify the slope and y-intercept. $$ f(x)=4-\frac{1}{2} x $$
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Solve the absolute value equation. $$|6 x-9|=0$$
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Find the slope-intercept form for the line satisfying the conditions. y-intercept \(-155,\) slope 5.6
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Solve the inequality symbolically. Express the solution set in set-builder or interval notation. $$ -5
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