Question

match each equation to its solution and to the story it describes
equations \t\t\t\t\t\tsolutions
a) $9x = 12 + 3$ \t\t\t\t\t1 $-4$
b) $3x = 8x + 10$ \t\t\t\t\t2 $\frac{15}{9}$
c) $4x + 5 = -11$ \t\t\t\t\t3 $-2$
d) $\frac{1}{2}x = 7x - 8$ \t\t\t\t\t4 $8$

the temperature is $-7$ since midnight the temperature rising and then rose 5 degrees. what was temperature at midnight?

jackie has 7 pink roses and some white roses. she gives all of them away, 5 roses to each of her 3 favorite teachers. how many white roses did she give away?

a musical instrument company reduced the time it takes for a worker to build a guitar. before the reduction it took 8 hours. now in 7 hours they can build 3 guitars. by how much did they reduce the time it takes to build each guitar?

a class splits its members into 5 groups for an activity. when 7 students have to leave early, there are only 8 students left to finish the activity. how many students were in each group?

Explanation:

Step1: Analyze the first sub-question (temperature problem)

Let the temperature at midnight be \( x \). It rose (let's assume rose by \( r \), but wait, the problem says "the temperature rising and then rose 5 degrees"? Wait, maybe typo. Wait, original: "The temperature is -7 since midnight the temperature rising and then rose 5 degrees. What was temperature at midnight?" Wait, maybe "The temperature now is -7. Since midnight, the temperature rose (or changed) and then rose 5 degrees? Wait, maybe rephrasing: Let \( x \) be midnight temp. Then \( x + \text{rise} + 5 = -7 \)? No, maybe "After rising some and then rising 5, now is -7. Wait, maybe the problem is: The current temperature is -7. Since midnight, the temperature rose (let's say the first rise is \( r \), but maybe the problem is simpler. Wait, maybe "The temperature is -7. Since midnight, the temperature rose and then rose 5 degrees. Wait, maybe the correct interpretation: Let \( x \) be midnight temp. Then \( x + 5 = -7 \)? No, that would be \( x = -12 \), but maybe. Wait, maybe the problem is: "The temperature now is -7. Since midnight, the temperature fell (or rose) and then rose 5 degrees. Wait, the text is unclear, but let's assume it's a linear equation. Let's suppose the problem is: "The temperature is -7. Since midnight, the temperature rose (let's say the change is \( x \) to \( x + 5 = -7 \))? No, maybe "The temperature at midnight was \( x \), then it rose (or changed) and now is -7 after rising 5. So \( x + 5 = -7 \), so \( x = -12 \). But maybe.

Step2: Analyze the second sub-question (roses)

Laila has 7 pink roses and \( w \) white roses. Total roses: \( 7 + w \). She gives 5 to each of 3 teachers, so total given: \( 5 \times 3 = 15 \). So \( 7 + w = 15 \), so \( w = 8 \).

Step3: Analyze the third sub-question (guitar company)

Before: time per guitar: 8 hours (for 1 guitar? Wait, "Before the reduction it took 8 hours. Now in 7 hours they can build 3 guitars." Wait, before: time to build 1 guitar: 8 hours. Now: time to build 3 guitars: 7 hours, so time per guitar: \( \frac{7}{3} \) hours? No, wait, "by how much did they reduce the time it takes to build each guitar?" So before: 8 hours per guitar. Now: \( \frac{7}{3} \)? No, wait, maybe "Before, to build 3 guitars, it took 8 hours? No, text: "Before the reduction it took 8 hours. Now in 7 hours they can build 3 guitars." So before: time per guitar: 8 hours (if building 1 guitar took 8 hours). Now: time per guitar: \( \frac{7}{3} \) hours? No, that can't be. Wait, maybe "Before, to build 1 guitar, 8 hours. Now, to build 3 guitars, 7 hours, so per guitar: \( \frac{7}{3} \approx 2.33 \), but reduction is \( 8 - \frac{7}{3} = \frac{24 - 7}{3} = \frac{17}{3} \approx 5.66 \), but that seems odd. Maybe "Before, building 3 guitars took 8 hours? No, text: "Before the reduction it took 8 hours. Now in 7 hours they can build 3 guitars." So before: time for 3 guitars: 8 hours? No, "it took 8 hours" for how many? Maybe "Before, to build 1 guitar, 8 hours. Now, in 7 hours, build 3 guitars, so time per guitar: \( \frac{7}{3} \) hours. Reduction: \( 8 - \frac{7}{3} = \frac{17}{3} \) hours per guitar.

Step4: Analyze the fourth sub-question (classroom groups)

A class is divided into 5 groups. 7 students leave, now 8 per group. So original total: \( 5g \), after 7 leave: \( 5g - 7 = 8 \times 5 \)? Wait, "When 7 students have to leave early, there are only 8 students left to finish the activity." Wait, no: "A classroom is numbered into 5 groups. You an activity. When 7 students have to leave early, there are only 8 students left to finish the activity. How many students were in each group?" Wait, total students: \( 5g \). After 7 leave, \( 5g - 7 = 8 \)? No, "only 8 students left" total? So \( 5g - 7 = 8 \), so \( 5g = 15 \), so \( g = 3 \). So each group had 3 students.

But the initial part of the image has equations and solutions (a) to (d) with equations like \( 9x = 12 - 3 \), \( 3x = 8 + 4 \), etc. Let's solve those:

(a) \( 9x = 12 - 3 \) → \( 9x = 9 \) → \( x = 1 \) (matches solution 1: -4? No, wait 12-3=9, 9x=9→x=1. But solution 1 is -4? Maybe typo.

(b) \( 3x = 8 + 4 \) → \( 3x = 12 \) → \( x = 4 \) (solution 2: \( \frac{12}{5} \)? No, 8+4=12, 3x=12→x=4.

(c) \( 4x + 5 = -11 \) → \( 4x = -16 \) → \( x = -4 \) (solution 3: 2? No, -11-5=-16, 4x=-16→x=-4.

(d) \( \frac{1}{2}x + 7 = -5 \) → \( \frac{1}{2}x = -12 \) → \( x = -24 \) (solution 4: 8? No, maybe equations are different.

Wait, maybe the first part is matching equations to solutions:

(a) \( 9x = 12 - 3 \) → \( 9x = 9 \) → \( x = 1 \) (solution 1: 1? But solution 1 is -4. Maybe equation is \( 9x = 12 + 3 \)? No.

(b) \( 3x = 8 - 4 \) → \( 3x = 4 \) → \( x = \frac{4}{3} \)? No, solution 2 is \( \frac{12}{5} \).

(c) \( 4x + 5 = -3 \) → \( 4x = -8 \) → \( x = -2 \) (solution 3: 2? No.

(d) \( \frac{1}{2}x + 7 = 11 \) → \( \frac{1}{2}x = 4 \) → \( x = 8 \) (solution 4: 8. Ah, here we go. So (d) \( \frac{1}{2}x + 7 = 11 \):

Step: \( \frac{1}{2}x = 11 - 7 = 4 \) → \( x = 4 \times 2 = 8 \). So (d) matches solution 4 (8).

So for the equation matching:

(a) Let's find equation for solution 1 (-4): \( 4x + 5 = -11 \) → \( 4x = -16 \) → \( x = -4 \). So (a) should be \( 4x + 5 = -11 \), solution -4.

(b) Solution 2: \( \frac{12}{5} \). Let's find equation: \( 5x = 12 \) → \( x = \frac{12}{5} \). So equation like \( 5x = 12 \), but in the image, (b) is \( 3x = 8 + 4 \)? No, maybe (b) is \( 5x = 12 \), so \( x = \frac{12}{5} \).

(c) Solution 3: 2. Equation: \( 9x = 18 \) → \( x = 2 \), but (c) is \( 4x + 5 = -3 \)? No, \( 9x = 18 \) → \( x = 2 \).

(d) Solution 4: 8. Equation: \( \frac{1}{2}x + 7 = 11 \) → \( x = 8 \), as above.

So the matching is:

(a) \( 4x + 5 = -11 \) → solution -4 (1)

(b) \( 5x = 12 \) → solution \( \frac{12}{5} \) (2)

(c) \( 9x = 18 \) → solution 2 (3)

(d) \( \frac{1}{2}x + 7 = 11 \) → solution 8 (4)

Now, for the sub-questions:

1. Temperature: Let's assume the problem is "The current temperature is -7. Since midnight, the temperature rose 5 degrees. What was the temperature at midnight?" So \( x + 5 = -7 \) → \( x = -12 \).

2. Roses: Laila has 7 pink and \( w \) white. Gives 5 to 3 teachers: \( 7 + w = 5 \times 3 = 15 \) → \( w = 8 \).

3. Guitar: Before, time per guitar: 8 hours (for 1 guitar). Now, in 7 hours, build 3 guitars, so time per guitar: \( \frac{7}{3} \) hours. Reduction: \( 8 - \frac{7}{3} = \frac{24 - 7}{3} = \frac{17}{3} \approx 5.67 \) hours? Wait, no, "by how much did they reduce the time it takes to build each guitar?" So original time per guitar: 8 hours (if building 1 guitar took 8 hours). New time per guitar: \( \frac{7}{3} \) hours (since 3 guitars in 7 hours: 7/3 per guitar). So reduction: \( 8 - \frac{7}{3} = \frac{17}{3} \) hours. But maybe the problem is "Before, to build 3 guitars, it took 8 hours. Now, to build 3 guitars, it takes 7 hours." Then time per guitar before: \( \frac{8}{3} \) hours, now: \( \frac{7}{3} \) hours. Reduction: \( \frac{8}{3} - \frac{7}{3} = \frac{1}{3} \) hours. Ah, that makes more sense. So "Before the reduction, it took 8 hours to build 3 guitars. Now, it takes 7 hours to build 3 guitars. By how much did they reduce the time per guitar?" So before: \( \frac{8}{3} \) hours per guitar, now: \( \frac{7}{3} \) hours per guitar. Reduction: \( \frac{8}{3} - \frac{7}{3} = \frac{1}{3} \) hours.

4. Classroom groups: Total students: \( 5g \) (5 groups, \( g \) per group). After 7 leave, \( 5g - 7 = 8 \) (total left is 8). So \( 5g = 15 \) → \( g = 3 \). So each group had 3 students.

But since the user's problem has multiple sub-questions, and the first part is equation matching, let's focus on the equation matching (the top part of the image).

For equation (d): \( \frac{1}{2}x + 7 = -5 \)? No, earlier we saw that \( \frac{1}{2}x + 7 = 11 \) gives \( x = 8 \), which matches solution 4 (8). So solving (d):

Step1: Subtract 7 from both sides

\( \frac{1}{2}x + 7 - 7 = 11 - 7 \) → \( \frac{1}{2}x = 4 \)

Step2: Multiply both sides by 2

\( \frac{1}{2}x \times 2 = 4 \times 2 \) → \( x = 8 \)

So the solution for (d) is 8, matching solution 4.

Answer:

For the equation \( \frac{1}{2}x + 7 = 11 \) (assuming (d) is this), the solution is \( \boldsymbol{8} \).

For the roses problem: White roses = \( \boldsymbol{8} \).

For the classroom groups: Students per group = \( \boldsymbol{3} \).

(Note: The image's text is partially unclear, but based on analysis, these are the solutions for the sub-questions.)