Solving a System of Linear Equations
Problem Statement
We need to find the coordinate point \((x, y)\) that satisfies both of the following linear equations:
$$ 3x + 4y + 5 = 0 $$ $$ y = x + 4 $$Equation 1: \( 3x + 4y + 5 = 0 \)
First, we rearrange the equation to isolate \(y\). This makes it easier to find points on the line.
$$ 4y = -3x - 5 $$ $$ y = \frac{-3x - 5}{4} $$Table of Values
| x | -3 | 1 | 5 |
|---|---|---|---|
| y | 1 | -2 | -5 |
| (x, y) | (-3, 1) | (1, -2) | (5, -5) |
Calculation Steps
For \(x = -3\):
\( y = \frac{-3(-3) - 5}{4} \)
\( y = \frac{9 - 5}{4} = \frac{4}{4} \)
\(\therefore y = 1 \)
For \(x = 1\):
\( y = \frac{-3(1) - 5}{4} \)
\( y = \frac{-3 - 5}{4} = \frac{-8}{4} \)
\(\therefore y = -2 \)
For \(x = 5\):
\( y = \frac{-3(5) - 5}{4} \)
\( y = \frac{-15 - 5}{4} = \frac{-20}{4} \)
\(\therefore y = -5 \)
Equation 2: \( y = x + 4 \)
This equation is already solved for \(y\), so we can directly calculate points.
Table of Values
| x | -3 | 0 | 1 |
|---|---|---|---|
| y | 1 | 4 | 5 |
| (x, y) | (-3, 1) | (0, 4) | (1, 5) |
Calculation Steps
For \(x = -3\):
\( y = (-3) + 4 \)
\(\therefore y = 1 \)
For \(x = 0\):
\( y = 0 + 4 \)
\(\therefore y = 4 \)
For \(x = 1\):
\( y = 1 + 4 \)
\(\therefore y = 5 \)
Graphical Solution
By plotting both lines on a graph, we can see they intersect at a single point. This point is the solution to the system of equations.
Conclusion
By comparing the tables of values and observing the point of intersection on the graph, we can see that the common point is \((-3, 1)\).