SOLUTION: Unit 1 Distance and Midpoint Formulas Geometry Basics ... - Free Printable

Problem Analysis and Solution

The image contains several problems related to solving systems of linear equations and inequalities. Let's solve each problem step by step.

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#### Exercise 1: Solve the system of equations

##### Problem 1
Solve the system:
\[
\begin{cases}
\frac{x + y}{2} = \frac{x - y}{3} \\
x + 2y = 4
\end{cases}
\]

Step 1: Simplify the first equation.
\[
\frac{x + y}{2} = \frac{x - y}{3}
\]
Cross-multiply to eliminate the fractions:
\[
3(x + y) = 2(x - y)
\]
Expand both sides:
\[
3x + 3y = 2x - 2y
\]
Rearrange terms to isolate \(x\) and \(y\):
\[
3x - 2x + 3y + 2y = 0 \implies x + 5y = 0 \quad \text{(Equation 1)}
\]

Step 2: Use the second equation:
\[
x + 2y = 4 \quad \text{(Equation 2)}
\]

Step 3: Solve the system of linear equations:
\[
\begin{cases}
x + 5y = 0 \\
x + 2y = 4
\end{cases}
\]
Subtract Equation 2 from Equation 1:
\[
(x + 5y) - (x + 2y) = 0 - 4
\]
Simplify:
\[
3y = -4 \implies y = -\frac{4}{3}
\]

Step 4: Substitute \(y = -\frac{4}{3}\) into Equation 2:
\[
x + 2\left(-\frac{4}{3}\right) = 4
\]
Simplify:
\[
x - \frac{8}{3} = 4 \implies x = 4 + \frac{8}{3} = \frac{12}{3} + \frac{8}{3} = \frac{20}{3}
\]

Solution:
\[
\boxed{\left( \frac{20}{3}, -\frac{4}{3} \right)}
\]

---

##### Problem 2
Solve the system:
\[
\begin{cases}
\frac{x + y}{2} = \frac{x - y}{3} \\
x + 2y = 4
\end{cases}
\]

This is the same system as Problem 1, so the solution is:
\[
\boxed{\left( \frac{20}{3}, -\frac{4}{3} \right)}
\]

---

##### Problem 3
Find the coordinates of the intersection of the lines:
- \(x + y = 4\)
- \(x - y = 0\)
- \(x + 2y = 4\)

Step 1: Solve the system of equations:
\[
\begin{cases}
x + y = 4 \\
x - y = 0
\end{cases}
\]
Add the two equations:
\[
(x + y) + (x - y) = 4 + 0 \implies 2x = 4 \implies x = 2
\]
Substitute \(x = 2\) into \(x - y = 0\):
\[
2 - y = 0 \implies y = 2
\]
So, the intersection point of \(x + y = 4\) and \(x - y = 0\) is \((2, 2)\).

Step 2: Verify if \((2, 2)\) satisfies \(x + 2y = 4\):
\[
2 + 2(2) = 2 + 4 = 6 \neq 4
\]
Thus, \((2, 2)\) does not lie on \(x + 2y = 4\).

Step 3: Solve the system:
\[
\begin{cases}
x + y = 4 \\
x + 2y = 4
\end{cases}
\]
Subtract the first equation from the second:
\[
(x + 2y) - (x + y) = 4 - 4 \implies y = 0
\]
Substitute \(y = 0\) into \(x + y = 4\):
\[
x + 0 = 4 \implies x = 4
\]
So, the intersection point of \(x + y = 4\) and \(x + 2y = 4\) is \((4, 0)\).

Step 4: Solve the system:
\[
\begin{cases}
x - y = 0 \\
x + 2y = 4
\end{cases}
\]
From \(x - y = 0\), we have \(x = y\). Substitute \(x = y\) into \(x + 2y = 4\):
\[
y + 2y = 4 \implies 3y = 4 \implies y = \frac{4}{3}
\]
Thus, \(x = \frac{4}{3}\). The intersection point is \(\left( \frac{4}{3}, \frac{4}{3} \right)\).

Final Answer for Problem 3:
\[
\boxed{(2, 2), (4, 0), \left( \frac{4}{3}, \frac{4}{3} \right)}
\]

---

#### Exercise 2: Solve the inequalities

##### Problem 1
Solve:
\[
\begin{cases}
2x + 3 > 5x - 6 \\
x - 3 < 0
\end{cases}
\]

Step 1: Solve the first inequality:
\[
2x + 3 > 5x - 6
\]
Rearrange terms:
\[
2x - 5x > -6 - 3 \implies -3x > -9 \implies x < 3
\]

Step 2: Solve the second inequality:
\[
x - 3 < 0 \implies x < 3
\]

Step 3: Combine the solutions:
Both inequalities give \(x < 3\). Therefore, the solution is:
\[
\boxed{x < 3}
\]

---

##### Problem 2
Solve:
\[
\begin{cases}
2x + 3 > 5x - 6 \\
x - 3 > 0
\end{cases}
\]

Step 1: Solve the first inequality (same as before):
\[
x < 3
\]

Step 2: Solve the second inequality:
\[
x - 3 > 0 \implies x > 3
\]

Step 3: Combine the solutions:
There is no overlap between \(x < 3\) and \(x > 3\). Therefore, the solution is:
\[
\boxed{\emptyset}
\]

---

##### Problem 3
Solve:
\[
\begin{cases}
6x + 2y > 16 \\
3x + y \leq 6
\end{cases}
\]

Step 1: Solve the first inequality:
\[
6x + 2y > 16 \implies 3x + y > 8
\]

Step 2: Solve the second inequality:
\[
3x + y \leq 6
\]

Step 3: Graph the inequalities:
- The line \(3x + y = 8\) is a boundary for \(3x + y > 8\).
- The line \(3x + y = 6\) is a boundary for \(3x + y \leq 6\).

The solution is the region where \(3x + y > 8\) and \(3x + y \leq 6\) overlap. However, these two regions do not overlap. Therefore, the solution is:
\[
\boxed{\emptyset}
\]

---

##### Problem 4
Solve:
\[
\begin{cases}
6x + 2y > 16 \\
3x + y \geq 6
\end{cases}
\]

Step 1: Solve the first inequality:
\[
6x + 2y > 16 \implies 3x + y > 8
\]

Step 2: Solve the second inequality:
\[
3x + y \geq 6
\]

Step 3: Graph the inequalities:
- The line \(3x + y = 8\) is a boundary for \(3x + y > 8\).
- The line \(3x + y = 6\) is a boundary for \(3x + y \geq 6\).

The solution is the region where \(3x + y > 8\) and \(3x + y \geq 6\) overlap. This region is above the line \(3x + y = 8\) and includes the region above and on the line \(3x + y = 6\).

Final Answer for Problem 4:
\[
\boxed{x > \frac{2}{3}, y > 2}
\]

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Final Answers

1. \(\boxed{\left( \frac{20}{3}, -\frac{4}{3} \right)}\)
2. \(\boxed{\left( \frac{20}{3}, -\frac{4}{3} \right)}\)
3. \(\boxed{(2, 2), (4, 0), \left( \frac{4}{3}, \frac{4}{3} \right)}\)
4. \(\boxed{x < 3}\)
5. \(\boxed{\emptyset}\)
6. \(\boxed{\emptyset}\)
7. \(\boxed{x > \frac{2}{3}, y > 2}\)