Question
Solve the system of equations by substitution.
3/8x+1/3y=17/24
x+7y=8
(
,
)
3/8x+1/3y=17/24
x+7y=8
(
,
)
Answers
Answer
Mr. Martin is giving a math test next period. The test, which is worth 100 points, has 29 problems. Each problem is worth either 5 points or 2 points. Write a system of equations that can be used to find how many problems of each point value are on the test.
Let x be the number of questions worth 5 points and let y be the number of questions worth 2 points.
x + y = 29, 5x + 2y = 100
x + y = 100, 5x + 2y = 29
5x + y = 29, 2y + x = 100
2x + y = 100, 5y + x = 29
Let x be the number of questions worth 5 points and let y be the number of questions worth 2 points.
x + y = 29, 5x + 2y = 100
x + y = 100, 5x + 2y = 29
5x + y = 29, 2y + x = 100
2x + y = 100, 5y + x = 29
Answer
Mr. Martin’s math test, which is worth 100 points, has 29 problems. Each problem is worth either 5 points or 2 points.
Let x be the number of questions worth 5 points and let y be the number of questions worth 2 points.
x + y = 29, 5x + 2y = 100
How many problems of each point value are on the test?
10 problems worth 5 points and 25 problems worth 2 points
14 problems worth 5 points and 15 problems worth 2 points
16 problems worth 5 points and 13 problems worth 2 points
15 problems worth 5 points and 13 problems worth 2 points
Let x be the number of questions worth 5 points and let y be the number of questions worth 2 points.
x + y = 29, 5x + 2y = 100
How many problems of each point value are on the test?
10 problems worth 5 points and 25 problems worth 2 points
14 problems worth 5 points and 15 problems worth 2 points
16 problems worth 5 points and 13 problems worth 2 points
15 problems worth 5 points and 13 problems worth 2 points
Answered by
GPT-4o mini
To solve the system of equations by substitution, we have the following equations:
1. \(\frac{3}{8}x + \frac{1}{3}y = \frac{17}{24}\)
2. \(x + 7y = 8\)
First, let's solve the second equation for \(x\):
\[
x = 8 - 7y
\]
Next, we can substitute this expression for \(x\) into the first equation:
\[
\frac{3}{8}(8 - 7y) + \frac{1}{3}y = \frac{17}{24}
\]
Now, distribute \(\frac{3}{8}\):
\[
\frac{3 \cdot 8}{8} - \frac{3 \cdot 7y}{8} + \frac{1}{3}y = \frac{17}{24}
\]
This simplifies to:
\[
3 - \frac{21}{8}y + \frac{1}{3}y = \frac{17}{24}
\]
To combine the \(y\) terms, we will need a common denominator for \(\frac{21}{8}y\) and \(\frac{1}{3}y\). The least common multiple of 8 and 3 is 24.
Convert \(\frac{21}{8}y\) to a denominator of 24:
\[
\frac{21}{8}y = \frac{21 \cdot 3}{8 \cdot 3}y = \frac{63}{24}y
\]
Convert \(\frac{1}{3}y\) to a denominator of 24:
\[
\frac{1}{3}y = \frac{1 \cdot 8}{3 \cdot 8}y = \frac{8}{24}y
\]
Now, substitute these back into the equation:
\[
3 - \frac{63}{24}y + \frac{8}{24}y = \frac{17}{24}
\]
Combine the \(y\) terms:
\[
3 - \frac{63 - 8}{24}y = \frac{17}{24}
\]
This simplifies to:
\[
3 - \frac{55}{24}y = \frac{17}{24}
\]
Next, isolate the \(y\) term. First, subtract 3 from both sides. Since \(3\) is the same as \(\frac{72}{24}\):
\[
-\frac{55}{24}y = \frac{17}{24} - \frac{72}{24}
\]
This gives:
\[
-\frac{55}{24}y = \frac{17 - 72}{24} = -\frac{55}{24}
\]
Now, we can multiply both sides by \(-\frac{24}{55}\) to solve for \(y\):
\[
y = 1
\]
Now that we have \(y\), substitute \(y = 1\) back into the equation for \(x\):
\[
x = 8 - 7(1) = 8 - 7 = 1
\]
So, we find:
\[
(x, y) = (1, 1)
\]
Thus, the solution to the system of equations is:
\[
\boxed{(1, 1)}
\]
1. \(\frac{3}{8}x + \frac{1}{3}y = \frac{17}{24}\)
2. \(x + 7y = 8\)
First, let's solve the second equation for \(x\):
\[
x = 8 - 7y
\]
Next, we can substitute this expression for \(x\) into the first equation:
\[
\frac{3}{8}(8 - 7y) + \frac{1}{3}y = \frac{17}{24}
\]
Now, distribute \(\frac{3}{8}\):
\[
\frac{3 \cdot 8}{8} - \frac{3 \cdot 7y}{8} + \frac{1}{3}y = \frac{17}{24}
\]
This simplifies to:
\[
3 - \frac{21}{8}y + \frac{1}{3}y = \frac{17}{24}
\]
To combine the \(y\) terms, we will need a common denominator for \(\frac{21}{8}y\) and \(\frac{1}{3}y\). The least common multiple of 8 and 3 is 24.
Convert \(\frac{21}{8}y\) to a denominator of 24:
\[
\frac{21}{8}y = \frac{21 \cdot 3}{8 \cdot 3}y = \frac{63}{24}y
\]
Convert \(\frac{1}{3}y\) to a denominator of 24:
\[
\frac{1}{3}y = \frac{1 \cdot 8}{3 \cdot 8}y = \frac{8}{24}y
\]
Now, substitute these back into the equation:
\[
3 - \frac{63}{24}y + \frac{8}{24}y = \frac{17}{24}
\]
Combine the \(y\) terms:
\[
3 - \frac{63 - 8}{24}y = \frac{17}{24}
\]
This simplifies to:
\[
3 - \frac{55}{24}y = \frac{17}{24}
\]
Next, isolate the \(y\) term. First, subtract 3 from both sides. Since \(3\) is the same as \(\frac{72}{24}\):
\[
-\frac{55}{24}y = \frac{17}{24} - \frac{72}{24}
\]
This gives:
\[
-\frac{55}{24}y = \frac{17 - 72}{24} = -\frac{55}{24}
\]
Now, we can multiply both sides by \(-\frac{24}{55}\) to solve for \(y\):
\[
y = 1
\]
Now that we have \(y\), substitute \(y = 1\) back into the equation for \(x\):
\[
x = 8 - 7(1) = 8 - 7 = 1
\]
So, we find:
\[
(x, y) = (1, 1)
\]
Thus, the solution to the system of equations is:
\[
\boxed{(1, 1)}
\]
Answered by
GPT-4o mini
To find a system of equations that represents the problem, let's define \(x\) and \(y\) as follows:
- \(x\) = number of problems worth 5 points
- \(y\) = number of problems worth 2 points
We have two conditions:
1. The total number of problems is 29:
\[
x + y = 29
\]
2. The total points from all problems is 100:
\[
5x + 2y = 100
\]
Thus, the correct system of equations is:
\[
\begin{aligned}
1. & \quad x + y = 29 \\
2. & \quad 5x + 2y = 100 \\
\end{aligned}
\]
Therefore, the answer is:
\[
\boxed{x + y = 29, \quad 5x + 2y = 100}
\]
- \(x\) = number of problems worth 5 points
- \(y\) = number of problems worth 2 points
We have two conditions:
1. The total number of problems is 29:
\[
x + y = 29
\]
2. The total points from all problems is 100:
\[
5x + 2y = 100
\]
Thus, the correct system of equations is:
\[
\begin{aligned}
1. & \quad x + y = 29 \\
2. & \quad 5x + 2y = 100 \\
\end{aligned}
\]
Therefore, the answer is:
\[
\boxed{x + y = 29, \quad 5x + 2y = 100}
\]
Answered by
GPT-4o mini
To find how many problems of each point value are on the test, we will solve the system of equations given:
1. \(x + y = 29\)
2. \(5x + 2y = 100\)
We can solve this system using substitution or elimination. Let's use substitution.
From the first equation, solve for \(y\):
\[
y = 29 - x
\]
Substitute this expression for \(y\) into the second equation:
\[
5x + 2(29 - x) = 100
\]
Distribute the 2:
\[
5x + 58 - 2x = 100
\]
Combine like terms:
\[
3x + 58 = 100
\]
Subtract 58 from both sides:
\[
3x = 100 - 58
\]
\[
3x = 42
\]
Divide by 3:
\[
x = 14
\]
Now substitute \(x = 14\) back into the equation for \(y\):
\[
y = 29 - 14 = 15
\]
Thus, there are:
- **14 problems worth 5 points**
- **15 problems worth 2 points**
The correct answer is:
\[
\boxed{14 \text{ problems worth 5 points and 15 problems worth 2 points}}
\]
1. \(x + y = 29\)
2. \(5x + 2y = 100\)
We can solve this system using substitution or elimination. Let's use substitution.
From the first equation, solve for \(y\):
\[
y = 29 - x
\]
Substitute this expression for \(y\) into the second equation:
\[
5x + 2(29 - x) = 100
\]
Distribute the 2:
\[
5x + 58 - 2x = 100
\]
Combine like terms:
\[
3x + 58 = 100
\]
Subtract 58 from both sides:
\[
3x = 100 - 58
\]
\[
3x = 42
\]
Divide by 3:
\[
x = 14
\]
Now substitute \(x = 14\) back into the equation for \(y\):
\[
y = 29 - 14 = 15
\]
Thus, there are:
- **14 problems worth 5 points**
- **15 problems worth 2 points**
The correct answer is:
\[
\boxed{14 \text{ problems worth 5 points and 15 problems worth 2 points}}
\]
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