https://file.aiquickdraw.com/mathbot/file/1724891686281sliqi5p0.jpg

Let's solve problems 6 to 10 using linear equations from the image.

### **Problem 6:**
\[
\begin{aligned}
2x + 3y &= 13 \\
4x - 2y &= 2 \\
\end{aligned}
\]

**Solution:**

1. Multiply the first equation by 2 to eliminate \(y\):
   \[
   4x + 6y = 26
   \]
2. Subtract the second equation from the modified first equation:
   \[
   (4x + 6y) - (4x - 2y) = 26 - 2 \\
   8y = 24
   \]
3. Solve for \(y\):
   \[
   y = 3
   \]
4. Substitute \(y = 3\) into the first equation to find \(x\):
   \[
   2x + 3(3) = 13 \\
   2x + 9 = 13 \\
   2x = 4 \\
   x = 2
   \]

**Solution:** \(x = 2\), \(y = 3\)

### **Problem 7:**
\[
\begin{aligned}
2x + 4y &= 10 \\
4x - y &= -5 \\
\end{aligned}
\]

**Solution:**

1. Multiply the second equation by 4 to eliminate \(y\):
   \[
   16x - 4y = -20
   \]
2. Add the first equation to the modified second equation:
   \[
   (2x + 4y) + (16x - 4y) = 10 + (-20) \\
   18x = -10
   \]
3. Solve for \(x\):
   \[
   x = -\frac{10}{18} = -\frac{5}{9}
   \]
4. Substitute \(x = -\frac{5}{9}\) into the first equation to find \(y\):
   \[
   2\left(-\frac{5}{9}\right) + 4y = 10 \\
   -\frac{10}{9} + 4y = 10 \\
   4y = 10 + \frac{10}{9} \\
   4y = \frac{100}{9} \\
   y = \frac{25}{9}
   \]

**Solution:** \(x = -\frac{5}{9}\), \(y = \frac{25}{9}\)

### **Problem 8:**
\[
\begin{aligned}
5x + 2y &= 15 \\
2x - y &= 5 \\
\end{aligned}
\]

**Solution:**

1. Multiply the second equation by 2 to eliminate \(y\):
   \[
   4x - 2y = 10
   \]
2. Add the first equation to the modified second equation:
   \[
   (5x + 2y) + (4x - 2y) = 15 + 10 \\
   9x = 25
   \]
3. Solve for \(x\):
   \[
   x = \frac{25}{9}
   \]
4. Substitute \(x = \frac{25}{9}\) into the second equation to find \(y\):
   \[
   2\left(\frac{25}{9}\right) - y = 5 \\
   \frac{50}{9} - y = 5 \\
   y = \frac{50}{9} - 5 \\
   y = \frac{50}{9} - \frac{45}{9} \\
   y = \frac{5}{9}
   \]

**Solution:** \(x = \frac{25}{9}\), \(y = \frac{5}{9}\)

### **Problem 9:**
\[
\begin{aligned}
3x - y &= 7 \\
2x + 2y &= 6 \\
\end{aligned}
\]

**Solution:**

1. Divide the second equation by 2:
   \[
   x + y = 3
   \]
2. Add this to the first equation to eliminate \(y\):
   \[
   (3x - y) + (x + y) = 7 + 3 \\
   4x = 10
   \]
3. Solve for \(x\):
   \[
   x = 2.5
   \]
4. Substitute \(x = 2.5\) into the modified second equation to find \(y\):
   \[
   2.5 + y = 3 \\
   y = 0.5
   \]

**Solution:** \(x = 2.5\), \(y = 0.5\)

### **Problem 10:**
\[
\begin{aligned}
x + 2y &= 11 \\
3x + 4y &= 21 \\
\end{aligned}
\]

**Solution:**

1. Multiply the first equation by 3 to eliminate \(x\):
   \[
   3x + 6y = 33
   \]
2. Subtract the second equation from the modified first equation:
   \[
   (3x + 6y) - (3x + 4y) = 33 - 21 \\
   2y = 12
   \]
3. Solve for \(y\):
   \[
   y = 6
   \]
4. Substitute \(y = 6\) into the first equation to find \(x\):
   \[
   x + 2(6) = 11 \\
   x + 12 = 11 \\
   x = -1
   \]

**Solution:** \(x = -1\), \(y = 6\)

Would you like more details or have any other questions?

Here are 5 related questions:
1. How do you determine when to add or subtract equations in a system?
2. What are the benefits of the substitution method versus elimination in solving systems?
3. How can you verify the solution of a system of linear equations?
4. What types of systems have no solution or infinite solutions?
5. How do you apply these methods to real-world problems?

**Tip:** When solving systems of equations, always simplify the equations as much as possible before applying the elimination or substitution method.

Explore how to solve systems of linear equations using the elimination and substitution methods with detailed step-by-step solutions. This page covers various problems involving linear equations, suitable for high school students.

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