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Difficulty: Very hardLinear Equations in One Variable

In the equation below, cc is a constant.

3c(2x−1)−2(x+4c)5=2x−3\frac{3c(2x - 1) - 2(x + 4c)}{5} = 2x - 3

If the equation has no solution for xx, what is the value of cc?

Answer: 2

Answer

The correct answer is 2.
To find the value of cc that results in no solution, we first clear the fraction by multiplying both sides of the equation by 5, giving 3c(2x−1)−2(x+4c)=10x−153c(2x - 1) - 2(x + 4c) = 10x - 15. Next, we expand the terms to get 6cx−3c−2x−8c=10x−156cx - 3c - 2x - 8c = 10x - 15, and group the xx terms and constant terms: (6c−2)x−11c=10x−15(6c - 2)x - 11c = 10x - 15. For a linear equation to have no solution, the coefficients of xx on both sides must be equal while the constant terms must be different. Setting the coefficients equal gives 6c−2=106c - 2 = 10, which solves to c=2c = 2. Checking the constant terms when c=2c = 2, we get −11(2)=−22-11(2) = -22 on the left and −15-15 on the right. Since −22≠−15-22 \neq -15, the equation has no solution, confirming c=2c = 2 is correct.

Step-by-Step Solution

1
Multiply both sides of the equation by 5 to eliminate the denominator.
3c(2x−1)−2(x+4c)=10x−153c(2x - 1) - 2(x + 4c) = 10x - 15
Clearing the denominator simplifies the equation into standard polynomial terms.
2
Distribute the terms on the left side of the equation.
6cx−3c−2x−8c=10x−156cx - 3c - 2x - 8c = 10x - 15
Applying the distributive property expands the expression so terms can be grouped.
3
Group the xx terms and constant terms on the left side.
(6c−2)x−11c=10x−15(6c - 2)x - 11c = 10x - 15
Putting the equation in the standard form Ax+B=Cx+DAx + B = Cx + D allows us to easily set up the conditions for no solution.
4
Set the coefficients of xx on both sides equal to each other.
6c−2=106c - 2 = 10, which simplifies to 6c=126c = 12, and thus c=2c = 2.
For the equation to have no solution, the variable terms on both sides must cancel each other out.
5
Verify that the constant terms are not equal when c=2c = 2.
The left-side constant is −11(2)=−22-11(2) = -22, and the right-side constant is −15-15. Since −22≠−15-22 \neq -15, the equation has no solution.
If the constant terms were equal, the equation would have infinitely many solutions instead of no solution.

Key Concept

Identifying conditions for a linear equation in one variable to have no solution.

Alternative Method

Instead of clearing the fraction first, write the left side of the equation as (6c−25)x−11c5(\frac{6c - 2}{5})x - \frac{11c}{5}. For there to be no solution, the coefficient of xx on the left side, 6c−25\frac{6c - 2}{5}, must equal the coefficient of xx on the right side, which is 2. Solving 6c−25=2\frac{6c - 2}{5} = 2 gives 6c−2=10  ⟹  c=26c - 2 = 10 \implies c = 2.
Estimated Time:2m 30s
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Topics

Linear Equations in One Variable

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