Answer:
Solution: x = 1, y = 5 or (1, 5).
Step-by-step explanation:
Given the systems of linear equations, y = 3x + 2 and y = -2x + 7:
I will demonstrate both solving for the solutions algebraically and graphically.
Algebraic Method:
Equation 1: y = 3x + 2
Equation 2: y = -2x + 7
Substitute the value of y = 3x + 2 from Equation 1 into Equation 2:
Equation 2: y = -2x + 7
⇒ 3x + 2 = -2x + 7
Add 2x to both sides:
3x + 2x + 2 = -2x + 2x + 7
5x + 2 = 7
Subtract 2 from both sides:
5x + 2 - 2 = 7 - 2
5x = 5
Divide both sides by 5 to isolate x:
[tex]\displaystyle\mathsf{\frac{5x}{5}\:=\:\frac{5}{5}}[/tex]
x = 1
Substitute the value of x into Equation 1 to solve for y:
Equation 1: y = 3x + 2
y = 3(1) + 2
y = 3 + 2
y = 5
Therefore, the solution to the given system is x = 1, y = 5 or (1, 5).
Verify:
Substitute the value of x = 1, and y = 5 into both equations:
Equation 1: y = 3x + 2
y = 3x + 2
5 = 3(1) + 2
5 = 5 (True statement)
Equation 2: y = -2x + 7
y = -2x + 7
5 = -2(1) + 7
5 = -2 + 7
5 = 5 (True statement).
Hence, we have the correct solution to both equations.
Graphing Method:
In order to solve for the solution graphically, start by plotting the y-intercepts of both equations.
Equation 1: y = 3x + 2
Slope = 3, y-intercept: (0, 2)
Equation 2: y = -2x + 7
Slope = -2, y-intercept: (0, 7)
After plotting the y-intercepts, use the slope (rise over run technique) of each equation to plot other points into the graph. In other words:
In Equation 1, the slope = 3 (rise 3 units, run 1 unit to the right).
For Equation 2, the slope = -2 (down 2 units, run 1 unit to the right).
Continue this process until you have at least 3 points to connect your line with.
The solution will be the point where the two lines intersect, which occurs at point (1, 5).
Attached is a screenshot of the graphed equations.
