Solved In Exercises 15-20, find the point of intersection of ... - Free Printable

Let's solve each of the problems from 15 to 20 step by step. We are asked to find the point of intersection of each pair of straight lines, which means solving the system of two equations for $ x $ and $ y $. The solution will be the point $ (x, y) $ that satisfies both equations.

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15.

$$
\begin{align*}
(1)\quad & y = 3x + 4 \\
(2)\quad & y = -2x + 14
\end{align*}
$$

Since both equations are solved for $ y $, set them equal:
$$
3x + 4 = -2x + 14
$$
Add $ 2x $ to both sides:
$$
5x + 4 = 14
$$
Subtract 4:
$$
5x = 10 \Rightarrow x = 2
$$
Now substitute $ x = 2 $ into equation (1):
$$
y = 3(2) + 4 = 6 + 4 = 10
$$
✔ Intersection point: $ (2, 10) $

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16.

$$
\begin{align*}
(1)\quad & y = -4x - 7 \\
(2)\quad & -y = 5x + 10
\end{align*}
$$

First, rewrite equation (2) to solve for $ y $:
$$
-y = 5x + 10 \Rightarrow y = -5x - 10
$$

Now set equations equal:
$$
-4x - 7 = -5x - 10
$$
Add $ 5x $ to both sides:
$$
x - 7 = -10
$$
Add 7:
$$
x = -3
$$
Substitute into equation (1):
$$
y = -4(-3) - 7 = 12 - 7 = 5
$$
✔ Intersection point: $ (-3, 5) $

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17.

$$
\begin{align*}
(1)\quad & 2x - 3y = 6 \\
(2)\quad & 3x + 6y = 16
\end{align*}
$$

We'll use elimination. Let’s eliminate one variable.

Multiply equation (1) by 2:
$$
4x - 6y = 12 \quad \text{(new eq1)}
$$
Now add to equation (2):
$$
(4x - 6y) + (3x + 6y) = 12 + 16 \\
7x = 28 \Rightarrow x = 4
$$

Substitute $ x = 4 $ into equation (1):
$$
2(4) - 3y = 6 \Rightarrow 8 - 3y = 6 \Rightarrow -3y = -2 \Rightarrow y = \frac{2}{3}
$$
✔ Intersection point: $ \left(4, \frac{2}{3}\right) $

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18.

$$
\begin{align*}
(1)\quad & 2x + 4y = 11 \\
(2)\quad & -5x + 3y = 5
\end{align*}
$$

Use elimination. Let's eliminate $ x $. Multiply equations to make coefficients of $ x $ match.

Multiply (1) by 5:
$$
10x + 20y = 55 \quad \text{(new eq1)}
$$

Multiply (2) by 2:
$$
-10x + 6y = 10 \quad \text{(new eq2)}
$$

Add them:
$$
(10x + 20y) + (-10x + 6y) = 55 + 10 \\
26y = 65 \Rightarrow y = \frac{65}{26} = \frac{5}{2}
$$

Now substitute $ y = \frac{5}{2} $ into equation (1):
$$
2x + 4\left(\frac{5}{2}\right) = 11 \Rightarrow 2x + 10 = 11 \Rightarrow 2x = 1 \Rightarrow x = \frac{1}{2}
$$

✔ Intersection point: $ \left(\frac{1}{2}, \frac{5}{2}\right) $

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19.

$$
\begin{align*}
(1)\quad & y = \frac{1}{4}x - 5 \\
(2)\quad & 2x - \frac{3}{2}y = 1
\end{align*}
$$

Substitute equation (1) into equation (2):

Replace $ y $ in (2):
$$
2x - \frac{3}{2}\left(\frac{1}{4}x - 5\right) = 1
$$

Distribute:
$$
2x - \frac{3}{2} \cdot \frac{1}{4}x + \frac{3}{2} \cdot 5 = 1 \\
2x - \frac{3}{8}x + \frac{15}{2} = 1
$$

Combine like terms:
$$
\left(2 - \frac{3}{8}\right)x + \frac{15}{2} = 1 \Rightarrow \frac{16}{8} - \frac{3}{8} = \frac{13}{8}x
$$
So:
$$
\frac{13}{8}x + \frac{15}{2} = 1
$$

Subtract $ \frac{15}{2} $:
$$
\frac{13}{8}x = 1 - \frac{15}{2} = \frac{2}{2} - \frac{15}{2} = -\frac{13}{2}
$$

Now solve for $ x $:
$$
x = -\frac{13}{2} \cdot \frac{8}{13} = -4
$$

Now plug $ x = -4 $ into equation (1):
$$
y = \frac{1}{4}(-4) - 5 = -1 - 5 = -6
$$

✔ Intersection point: $ (-4, -6) $

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20.

$$
\begin{align*}
(1)\quad & y = \frac{2}{3}x - 4 \\
(2)\quad & x + 3y + 3 = 0
\end{align*}
$$

Substitute equation (1) into equation (2):

Replace $ y $:
$$
x + 3\left(\frac{2}{3}x - 4\right) + 3 = 0
$$

Simplify:
$$
x + 2x - 12 + 3 = 0 \Rightarrow 3x - 9 = 0 \Rightarrow 3x = 9 \Rightarrow x = 3
$$

Now plug into equation (1):
$$
y = \frac{2}{3}(3) - 4 = 2 - 4 = -2
$$

✔ Intersection point: $ (3, -2) $

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✔ Final Answers:

| Problem | Point of Intersection |
|--------|------------------------|
| 15 | $ (2, 10) $ |
| 16 | $ (-3, 5) $ |
| 17 | $ \left(4, \frac{2}{3}\right) $ |
| 18 | $ \left(\frac{1}{2}, \frac{5}{2}\right) $ |
| 19 | $ (-4, -6) $ |
| 20 | $ (3, -2) $ |

Let me know if you'd like a graph or explanation for any specific problem!