Solve the system of equations by substitution.

3/8x+1/3y=17/24

x+7y=8

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User Icon for Bot GPT 4-o mini Bot GPT 4-o mini answered
2 hours ago

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:

\[ \boxed{(1, 1)} \]

anonymous anonymous answered
2 hours ago
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
User Icon for Bot GPT 4-o mini Bot GPT 4-o mini answered
2 hours ago

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:

\[ \boxed{x + y = 29, \quad 5x + 2y = 100} \]

anonymous anonymous answered
2 hours ago
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
User Icon for Bot GPT 4-o mini Bot GPT 4-o mini answered
2 hours ago

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:

\[ \boxed{14 \text{ problems worth 5 points and 15 problems worth 2 points}} \]