Question
system of equations. \( x \) to the corresponding \[ \begin{array}{l} x+3 y=6 \\ 2 x-3 y=12 \end{array} \] \( x+2 y=4 \) \( -3 x+y=2 \) \( \square \) \( x+3 y=12 \) \( 2 x+y=39 \) \( \square \) \( 4 x-5 y=15 \) \( 2 x-5 y=11 \) \( \square \) \( 4 x+2 y=-2 \) \[ x+3 y=12 \] \( \square \)
Ask by Carter Little. in the United States
Jan 22,2025
Real Tutor Solution
Tutor-Verified Answer
Answer
Here are the solutions to the systems of equations:
1. \( (x, y) = (6, 0) \)
2. \( (x, y) = (0, 2) \)
3. \( (x, y) = (21, -3) \)
4. \( (x, y) = \left(2, -\frac{7}{5}\right) \)
5. \( (x, y) = (-3, 5) \)
Solution
Sure, let's solve each of the given systems of equations step by step. We'll find the values of \( x \) and \( y \) that satisfy each system.
---
### **System 1**
\[
\begin{cases}
x + 3y = 6 \\
2x - 3y = 12
\end{cases}
\]
**Solution:**
1. **Add the two equations to eliminate \( y \):**
\[
(x + 2x) + (3y - 3y) = 6 + 12 \\
3x = 18 \\
x = 6
\]
2. **Substitute \( x = 6 \) into the first equation:**
\[
6 + 3y = 6 \\
3y = 0 \\
y = 0
\]
**Solution:** \( \boxed{(x, y) = (6,\ 0)} \)
---
### **System 2**
\[
\begin{cases}
x + 2y = 4 \\
-3x + y = 2
\end{cases}
\]
**Solution:**
1. **Express \( x \) from the first equation:**
\[
x = 4 - 2y
\]
2. **Substitute \( x = 4 - 2y \) into the second equation:**
\[
-3(4 - 2y) + y = 2 \\
-12 + 6y + y = 2 \\
7y = 14 \\
y = 2
\]
3. **Find \( x \):**
\[
x = 4 - 2(2) = 0
\]
**Solution:** \( \boxed{(x, y) = (0,\ 2)} \)
---
### **System 3**
\[
\begin{cases}
x + 3y = 12 \\
2x + y = 39
\end{cases}
\]
**Solution:**
1. **Express \( x \) from the first equation:**
\[
x = 12 - 3y
\]
2. **Substitute \( x = 12 - 3y \) into the second equation:**
\[
2(12 - 3y) + y = 39 \\
24 - 6y + y = 39 \\
-5y = 15 \\
y = -3
\]
3. **Find \( x \):**
\[
x = 12 - 3(-3) = 12 + 9 = 21
\]
**Solution:** \( \boxed{(x, y) = (21,\ -3)} \)
---
### **System 4**
\[
\begin{cases}
4x - 5y = 15 \\
2x - 5y = 11
\end{cases}
\]
**Solution:**
1. **Subtract the second equation from the first to eliminate \( y \):**
\[
(4x - 2x) + (-5y + 5y) = 15 - 11 \\
2x = 4 \\
x = 2
\]
2. **Substitute \( x = 2 \) into the second equation:**
\[
2(2) - 5y = 11 \\
4 - 5y = 11 \\
-5y = 7 \\
y = -\frac{7}{5} \quad \text{or} \quad y = -1.4
\]
**Solution:** \( \boxed{(x, y) = \left(2,\ -\frac{7}{5}\right)} \)
---
### **System 5**
\[
\begin{cases}
4x + 2y = -2 \\
x + 3y = 12
\end{cases}
\]
**Solution:**
1. **Express \( x \) from the second equation:**
\[
x = 12 - 3y
\]
2. **Substitute \( x = 12 - 3y \) into the first equation:**
\[
4(12 - 3y) + 2y = -2 \\
48 - 12y + 2y = -2 \\
-10y = -50 \\
y = 5
\]
3. **Find \( x \):**
\[
x = 12 - 3(5) = 12 - 15 = -3
\]
**Solution:** \( \boxed{(x, y) = (-3,\ 5)} \)
---
Feel free to reach out if you need further assistance with these systems or any other questions!
Reviewed and approved by the UpStudy tutoring team
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Simplify this solution Bonus Knowledge
To solve a system of equations, one can use methods such as substitution or elimination. For example, with the first set \( x + 3y = 6 \) and \( 2x - 3y = 12 \), you can isolate \( x \) in the first equation and substitute into the second to find the values of \( x \) and \( y \). In real-world terms, systems of equations are super handy! For instance, if you run a coffee shop, you might want to know how many cups of different drinks to make based on sales forecasts—the equations can represent those relationships. Solving them helps manage resources effectively and optimize profits!
