Linear Equations in One Variable Flashcards

For what value of k does the equation 2(3x + k) = 6x − 10 have infinitely many solutions?

Answer: -5

Distributing the left side gives 6x + 2k = 6x − 10. The x-terms cancel from both sides, leaving 2k = −10, so k = −5. When k = −5, both sides are identical (6x − 10 = 6x − 10), which is true for every value of x — hence infinitely many solutions. Any other value of k makes the constants unequal, producing no solution.

How many solutions does the equation (2(x + 3))/3 − (x − 1)/2 = (x + 5)/6 have?

Answer: No solution

Multiply every term by the LCD of 6: 4(x + 3) − 3(x − 1) = (x + 5). Expanding: 4x + 12 − 3x + 3 = x + 5, which simplifies to x + 15 = x + 5. Subtracting x from both sides gives 15 = 5, a contradiction. Since no value of x can make 15 equal 5, the equation has no solution.

Three consecutive integers have the property that their sum equals 5 more than twice the largest integer. What is the middle integer?

Answer: 7

Let the middle integer be n, so the three consecutive integers are n − 1, n, and n + 1. Setting up the equation: (n − 1) + n + (n + 1) = 2(n + 1) + 5. The left side simplifies to 3n, and the right side to 2n + 7. Solving 3n = 2n + 7 gives n = 7. The three integers are 6, 7, and 8, and their sum (21) equals 2(8) + 5 = 21. ✓

A store prices a jacket at 40% above its wholesale cost. For a clearance sale, the store then reduces the marked-up price by 25%. If the final sale price is $63, what was the original wholesale cost of the jacket?

Answer: $60

Let c = wholesale cost. After a 40% markup, the price is 1.4c. After a 25% reduction: 0.75 × 1.4c = 1.05c. Setting 1.05c = 63 and solving: c = 63 ÷ 1.05 = $60. A common error is subtracting the two percentages (40% − 25% = 15% markup) and solving 1.15c = 63, which gives the wrong answer.

For what value of a does the equation a(x + 1) = 4x + 3 have no solution?

Answer: 4

Expanding: ax + a = 4x + 3. Rearranging: (a − 4)x = 3 − a. If a ≠ 4, there is exactly one solution: x = (3 − a)/(a − 4). If a = 4, the left side becomes 0 · x = 0, and the right side becomes 3 − 4 = −1, giving 0 = −1, a contradiction. So a = 4 produces no solution. Note: if the right side had also been 0 (e.g., 3 − a = 0, meaning a = 3), we'd have infinitely many solutions instead.

The sum of three numbers is 100. The second number is 3 less than twice the first, and the third number is 7 more than three times the first. What is the largest of the three numbers?

Answer: 55

Let the first number be x. Then the second is 2x − 3, and the third is 3x + 7. The equation is x + (2x − 3) + (3x + 7) = 100, which simplifies to 6x + 4 = 100, giving 6x = 96 and x = 16. The three numbers are 16, 2(16) − 3 = 29, and 3(16) + 7 = 55. Their sum is 16 + 29 + 55 = 100 ✓. The largest is 55.