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solve the equation 3/5x-10=18-4x-1

Simplify using the distributive property. Which number can be distributed across two terms inside parentheses?___

-10
3/5
x

??? which one

11 answers

To solve the equation \( \frac{3}{5}x - 10 = 18 - 4x - 1 \), first let's simplify the right side of the equation.

  1. Combine like terms on the right: \[ 18 - 1 = 17 \] So, the equation becomes: \[ \frac{3}{5}x - 10 = 17 - 4x \]

Next, we'll move all terms involving \( x \) to one side and constants to the other side.

  1. Add \( 4x \) to both sides: \[ \frac{3}{5}x + 4x - 10 = 17 \]

  2. To combine \( \frac{3}{5}x \) and \( 4x \), we can convert \( 4x \) into a fraction: \[ 4x = \frac{20}{5}x \] Then, combine: \[ \frac{3}{5}x + \frac{20}{5}x = \frac{23}{5}x \]

Now the equation looks like this: \[ \frac{23}{5}x - 10 = 17 \]

  1. Add \( 10 \) to both sides: \[ \frac{23}{5}x = 27 \]

  2. To isolate \( x \), multiply both sides by the reciprocal of \( \frac{23}{5} \) (which is \( \frac{5}{23} \)): \[ x = 27 \cdot \frac{5}{23} \] \[ x = \frac{135}{23} \]

Now, regarding the original question about which number can be distributed across two terms inside parentheses:

  • The options are \( -10 \), \( \frac{3}{5} \), or \( x \).

In the original equation, if we had any parentheses (which we do not currently), typically, you would distribute a coefficient (like \( \frac{3}{5} \) or a constant). However, in this equation, we initially do not have parentheses to distribute over.

If we were to apply it to a situation where one of these numbers could "distribute" in a hypothetical scenario, \( \frac{3}{5} \) would typically be the number in front of a term in parentheses, but in the context of the current equation, it is not applicable because there are no parentheses.

So, there is no clear application for distribution in this equation given there are no parentheses present. Therefore, the answer to your question is that none of the numbers you listed can be distributed because there are no terms within parentheses in the equation provided.

its 3/5

now we have 3/5x-6=18-4x-1

Combine like terms that are on the same side of the equation. Which terms can be combined?

Let's simplify the new equation you provided:

\[ \frac{3}{5}x - 6 = 18 - 4x - 1 \]

First, combine like terms on the right side:

  1. Combine the constants \(18\) and \(-1\): \[ 18 - 1 = 17 \] So, the equation now reads: \[ \frac{3}{5}x - 6 = 17 - 4x \]

Now, we can examine the left side of the equation:

  • On the left side, the only terms are \(\frac{3}{5}x\) and \(-6\). These cannot be combined because one is a variable term and the other is a constant.

On the right side:

  • The terms \(17\) and \(-4x\) cannot be combined either, as one is a constant and one is a variable term.

In summary:

  • The only like terms that can be combined based on the initial arrangement are the constant terms on the right side: \(18\) and \(-1\).

So, to answer your question: The terms that can be combined are \(18\) and \(-1\) on the right side of the equation.

The simplified equation after this step is: \[ \frac{3}{5}x - 6 = 17 - 4x \]

okay guys its
3/5
18 and -1
adding 6
adding 4x
5/23

solve the equation

-10-2x=3x+15

To solve the equation:

\[ -10 - 2x = 3x + 15 \]

we'll start by moving all terms involving \(x\) to one side and constant terms to the other side.

  1. First, let's add \(2x\) to both sides of the equation to get the \(x\) terms on one side:

\[ -10 = 3x + 2x + 15 \]

This simplifies to:

\[ -10 = 5x + 15 \]

  1. Next, we'll subtract \(15\) from both sides to isolate the term with \(x\):

\[ -10 - 15 = 5x \]

This simplifies to:

\[ -25 = 5x \]

  1. Now, divide both sides by \(5\) to solve for \(x\):

\[ x = \frac{-25}{5} \]

This further simplifies to:

\[ x = -5 \]

Thus, the solution to the equation \(-10 - 2x = 3x + 15\) is:

\[ \boxed{-5} \]

guys you put
distribute
combine terms
add 2x
subtract 15
divide 5
and there ya go boo

Sort each equation according to whether it has one solution, infinitely many solutions, or no solution.

5(x - 2) = 5x - 7

-3(x - 4) = -3x + 12

4(x + 1) = 3x + 4

-2(x - 3) = 2x - 6

6(x + 5) = 6x + 11

each section: One Solution

Infinitely Many Solutions

No Solution

Let's analyze each equation you provided to determine whether they have one solution, infinitely many solutions, or no solution. I'll solve each equation step by step.

1. \( 5(x - 2) = 5x - 7 \)

Distributing on the left side: \[ 5x - 10 = 5x - 7 \]

Subtract \(5x\) from both sides: \[ -10 = -7 \] This is false, meaning there are no values of \(x\) that will satisfy the equation.

Conclusion: No Solution

2. \( -3(x - 4) = -3x + 12 \)

Distributing on the left side: \[ -3x + 12 = -3x + 12 \]

Add \(3x\) to both sides: \[ 12 = 12 \] This is a true statement, meaning any value of \(x\) will satisfy the equation.

Conclusion: Infinitely Many Solutions

3. \( 4(x + 1) = 3x + 4 \)

Distributing on the left side: \[ 4x + 4 = 3x + 4 \]

Subtract \(3x\) from both sides: \[ 4x - 3x + 4 = 4 \] This simplifies to: \[ x + 4 = 4 \]

Subtract \(4\) from both sides: \[ x = 0 \]

Conclusion: One Solution \(x = 0\)

4. \( -2(x - 3) = 2x - 6 \)

Distributing on the left side: \[ -2x + 6 = 2x - 6 \]

Add \(2x\) to both sides: \[ 6 = 4x - 6 \]

Add \(6\) to both sides: \[ 12 = 4x \]

Divide by \(4\): \[ x = 3 \]

Conclusion: One Solution \(x = 3\)

5. \( 6(x + 5) = 6x + 11 \)

Distributing on the left side: \[ 6x + 30 = 6x + 11 \]

Subtract \(6x\) from both sides: \[ 30 = 11 \] This is false, meaning there are no values of \(x\) that will satisfy the equation.

Conclusion: No Solution

Summary

  • One Solution:

    • \(4(x + 1) = 3x + 4\) (Solution: \(x = 0\))
    • \(-2(x - 3) = 2x - 6\) (Solution: \(x = 3\))

  • Infinitely Many Solutions:

    • \(-3(x - 4) = -3x + 12\)

  • No Solution:

    • \(5(x - 2) = 5x - 7\)
    • \(6(x + 5) = 6x + 11\)

This categorization should help clarify the types of solutions for each equation!

Micah solves a linear equation and concludes that x = 0 is the solution. His work is shown below.

(1 – 3x) = 4(– + 2)
4 lines of math. The first line is, StartFraction 5 Over 6 EndFraction left-parenthesis 1 minus 3 x right-parenthesis equals 4 left-parenthesis negative StartFraction 5 x Over EndFraction plus 2 right-parenthesis. The second line is, StartFraction 5 Over 6 EndFraction minus StartFraction 5x Over 2 EndFraction equals StartFraction 5x Over 2 EndFraction plus 8. The third line in plus StartFraction 5x Over 2 EndFraction and StartFraction 5x Over 2 EndFraction on both sides of the equal sign. The fourth line is 0 equals x.

0 = x
Which statement is true about Micah’s solution?

Micah’s solution is wrong. There are no values of x that make the statement true.
Micah’s solution is correct, and the value of x that makes the statement true is 0.
Micah should have divided by StartFraction 5 Over 2 EndFraction.
Micah should have subtracted StartFraction 5 Over 2 EndFraction.

Let’s analyze Micah's work step by step to determine if the conclusions he drew about the solution being \( x = 0 \) are correct.

The equation Micah started with is:

\[ \frac{5}{6}(1 - 3x) = 4\left(-\frac{5x}{2} + 2\right) \]

  1. First Line Written by Micah: \[ \frac{5}{6}(1 - 3x) = 4\left(-\frac{5x}{2} + 2\right) \]

  2. Second Line: We’ll simplify both sides:

    • Left Side: \[ \frac{5}{6}(1 - 3x) = \frac{5}{6} - \frac{15x}{6} = \frac{5}{6} - \frac{5x}{2} \]
    • Right Side: \[ 4\left(-\frac{5x}{2} + 2\right) = -10x + 8 \]

    Therefore, Micah's second line seems incorrect based on the current logical flow of math.

  3. Correcting the Equation: The equation becomes: \[ \frac{5}{6} - \frac{5x}{2} = -10x + 8 \]

  4. Moving Terms Involving \(x\): Adding \(10x\) to both sides, \[ \frac{5}{6} + 10x - \frac{5x}{2} = 8 \]

  5. Combining x Terms: To combine \(10x\) and \(-\frac{5x}{2}\):

    • Convert \(10x\) to a fraction: \(10x = \frac{20x}{2}\)
    • Combine: \[ \frac{20x}{2} - \frac{5x}{2} = \frac{15x}{2} \]

Thus, you then have: \[ \frac{5}{6} + \frac{15x}{2} = 8 \]

  1. Move the constant to the other side: \[ \frac{15x}{2} = 8 - \frac{5}{6} \]

To simplify \(8 - \frac{5}{6}\), first convert \(8\) to a fraction: \[ 8 = \frac{48}{6} \Rightarrow \frac{48}{6} - \frac{5}{6} = \frac{43}{6} \]

So now we have: \[ \frac{15x}{2} = \frac{43}{6} \]

  1. Solve for \(x\): Multiply both sides by the reciprocal of \(\frac{15}{2}\): \[ x = \frac{43}{6} \cdot \frac{2}{15} = \frac{86}{90} = \frac{43}{45} \]

Conclusion:

Based on this analysis, Micah concluded \(x = 0\) incorrectly. The actual value of \(x\) is \(\frac{43}{45}\).

Now we examine the provided statements:

  1. Micah’s solution is wrong. There are no values of \(x\) that make the statement true. (This is false—there is a value of \(x\).)
  2. Micah’s solution is correct, and the value of \(x\) that makes the statement true is 0. (This is false; \(x\) is \(\frac{43}{45}\).)
  3. Micah should have divided by \(\frac{5}{2}\). (This is not applicable in the steps provided.)
  4. Micah should have subtracted \(\frac{5}{2}\). (While this might not have been the correct action in this context, he didn’t perform it anyway.)

The only true statement is: Micah’s solution is wrong. There are no values of \(x\) that make the statement true. (Based on how we interpret zero solutions, but there is indeed one.) As written, option 1 has contradictory truth. Thus, this phrase is mostly false due to its seriousness on values.

In the strictest interpretation, Micah’s work ultimately should be viewed as wrong given he reached the incorrect conclusion.

Raj and Aditi like to play chess against each other. Aditi has won 12 of the 20 games so far. They decide to play more games.

How many more games will Aditi have to win to have a 70% winning record?

12 + x = 0.70(20 + x)
x = x equals StartFraction 20 Over 3 EndFraction is approximately 6.67. ≈ 6.67
She will have to win 7 more games to have a 70% winning record.

How many more games will Aditi have to win to have an 80% winning record?

The linear equation that models this is
.

Aditi would have to win
more games to have an 80% winning record.

To find out how many more games Aditi needs to win to achieve an 80% winning record, we start with her current record and set up an equation.

Currently, Aditi has won 12 games out of 20. If she wins \( x \) more games, the total number of games played will be \( 20 + x \), and the total number of games she has won will be \( 12 + x \).

We want her winning percentage to be 80%, which can be expressed as:

\[ \frac{12 + x}{20 + x} = 0.80 \]

Now, we can solve this equation step by step.

  1. Cross-multiply to eliminate the fraction: \[ 12 + x = 0.80(20 + x) \]

  2. Distribute the \(0.80\) on the right side: \[ 12 + x = 16 + 0.80x \]

  3. Now, isolate \(x\). Start by subtracting \(0.80x\) from both sides: \[ 12 + x - 0.80x = 16 \] \[ 12 + 0.20x = 16 \]

  4. Next, subtract \(12\) from both sides: \[ 0.20x = 16 - 12 \] \[ 0.20x = 4 \]

  5. Finally, divide both sides by \(0.20\) to solve for \(x\): \[ x = \frac{4}{0.20} = 20 \]

Therefore, Aditi will have to win 20 more games to achieve an 80% winning record.

Summary:

  • Aditi needs to win 20 more games to have an 80% winning record.