Question
Solve the system of linear equations by graphing.
2x + 3y = 16.9
5x = y + 7.4
What is the solution to the system of linear equations? Round to the nearest tenth as needed.
2x + 3y = 16.9
5x = y + 7.4
What is the solution to the system of linear equations? Round to the nearest tenth as needed.
Answers
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Answered by
GPT-4o mini
To solve the system of linear equations by graphing, we need to rearrange both equations into slope-intercept form (y = mx + b) to find their intersections.
### Equation 1: \(2x + 3y = 16.9\)
Let's solve for \(y\):
1. Subtract \(2x\) from both sides:
\[
3y = 16.9 - 2x
\]
2. Divide by 3:
\[
y = \frac{16.9 - 2x}{3}
\]
\[
y = \frac{16.9}{3} - \frac{2}{3}x
\]
\[
y \approx 5.6333 - 0.6667x
\]
### Equation 2: \(5x = y + 7.4\)
Let's also solve this for \(y\):
1. Subtract \(7.4\) from both sides:
\[
y = 5x - 7.4
\]
### Now we have the two equations:
1. \(y \approx 5.6333 - 0.6667x\)
2. \(y = 5x - 7.4\)
### Graphing the two equations:
1. **Equation 1** has a y-intercept of approximately \(5.6333\) and a slope of \(-0.6667\).
2. **Equation 2** has a y-intercept of \(-7.4\) and a slope of \(5\).
### Finding points to graph:
#### For Equation 1:
- When \(x = 0\):
\[
y \approx 5.6333
\]
- When \(x = 5\):
\[
y \approx 5.6333 - 0.6667(5) \approx 5.6333 - 3.3335 \approx 2.2998
\]
#### For Equation 2:
- When \(x = 0\):
\[
y = 5(0) - 7.4 = -7.4
\]
- When \(x = 3\):
\[
y = 5(3) - 7.4 = 15 - 7.4 = 7.6
\]
### Graphing these points:
Plotting the points and drawing the lines for both functions, we can find their intersection point graphically.
### Solving Algebraically:
To find the intersection point algebraically:
Set the equations equal to each other:
\[
5x - 7.4 = 5.6333 - 0.6667x
\]
Combine like terms:
\[
5x + 0.6667x = 5.6333 + 7.4
\]
\[
5.6667x = 13.0333
\]
Now, solve for \(x\):
\[
x \approx \frac{13.0333}{5.6667} \approx 2.303
\]
Rounded to the nearest tenth, \(x \approx 2.3\).
Substituting \(x\) back into one of the equations to find \(y\):
Using \(y = 5x - 7.4\):
\[
y \approx 5(2.303) - 7.4 \approx 11.515 - 7.4 \approx 4.115
\]
Rounded to the nearest tenth, \(y \approx 4.1\).
### Solution:
The solution to the system of linear equations is:
\[
(x, y) \approx (2.3, 4.1)
\]
### Equation 1: \(2x + 3y = 16.9\)
Let's solve for \(y\):
1. Subtract \(2x\) from both sides:
\[
3y = 16.9 - 2x
\]
2. Divide by 3:
\[
y = \frac{16.9 - 2x}{3}
\]
\[
y = \frac{16.9}{3} - \frac{2}{3}x
\]
\[
y \approx 5.6333 - 0.6667x
\]
### Equation 2: \(5x = y + 7.4\)
Let's also solve this for \(y\):
1. Subtract \(7.4\) from both sides:
\[
y = 5x - 7.4
\]
### Now we have the two equations:
1. \(y \approx 5.6333 - 0.6667x\)
2. \(y = 5x - 7.4\)
### Graphing the two equations:
1. **Equation 1** has a y-intercept of approximately \(5.6333\) and a slope of \(-0.6667\).
2. **Equation 2** has a y-intercept of \(-7.4\) and a slope of \(5\).
### Finding points to graph:
#### For Equation 1:
- When \(x = 0\):
\[
y \approx 5.6333
\]
- When \(x = 5\):
\[
y \approx 5.6333 - 0.6667(5) \approx 5.6333 - 3.3335 \approx 2.2998
\]
#### For Equation 2:
- When \(x = 0\):
\[
y = 5(0) - 7.4 = -7.4
\]
- When \(x = 3\):
\[
y = 5(3) - 7.4 = 15 - 7.4 = 7.6
\]
### Graphing these points:
Plotting the points and drawing the lines for both functions, we can find their intersection point graphically.
### Solving Algebraically:
To find the intersection point algebraically:
Set the equations equal to each other:
\[
5x - 7.4 = 5.6333 - 0.6667x
\]
Combine like terms:
\[
5x + 0.6667x = 5.6333 + 7.4
\]
\[
5.6667x = 13.0333
\]
Now, solve for \(x\):
\[
x \approx \frac{13.0333}{5.6667} \approx 2.303
\]
Rounded to the nearest tenth, \(x \approx 2.3\).
Substituting \(x\) back into one of the equations to find \(y\):
Using \(y = 5x - 7.4\):
\[
y \approx 5(2.303) - 7.4 \approx 11.515 - 7.4 \approx 4.115
\]
Rounded to the nearest tenth, \(y \approx 4.1\).
### Solution:
The solution to the system of linear equations is:
\[
(x, y) \approx (2.3, 4.1)
\]
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