A system of linear equations includes the line that is created by the equation y = x+ 3, graphed below, and the line through the
points (3, 1) and (4, 3).
On a coordinate plane, a line goes through (0, 3) and (2, 5).
What is the solution to the system of equations?
On a coordinate plane, a line goes through (0, 3) and (2, 5).
What is the solution to the system of equations?
1 answer:
Answer:
(8, 11)
Step-by-step explanation:
The 2-point form of the equation of a line is useful for the one where only points are given.
y = (y2 -y1)/(x2 -x1)(x -x1) +y1
y = (3 -1)/(4 -3)(x -3) +1
y = 2(x -3) +1
We can equate this to the expression for y in the given equation.
x +3 = 2(x -3) +1
x +3 = 2x -5 . . . . . . eliminate parentheses
8 = x . . . . . . . . . . . . add 5-x
Using the given equation, we can find y:
y = x +3 = 8 +3
y = 11
The solution to the system is (x, y) = (8, 11).
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Problem 4
<h3>Answer:</h3>------------------
Work Shown:
The left line goes through (-2,0) and (0,4)
The slope of this line is
m = (y2-y1)/(x2-x1)
m = (4-0)/(0-(-2))
m = (4-0)/(0+2)
m = 4/2
m = 2
The y intercept is b = 4
Since m = 2 and b = 4, this means y = mx+b turns into y = 2x+4. This portion is only done when x < 1. Note the open circle at the endpoint of this portion. So we do not include x = 1 as part of this piece.
---
The line on the right side goes through (1,-2) and (2,-1)
Slope
m = (y2-y1)/(x2-x1)
m = (-1-(-2))/(2-1)
m = (-1+2)/(2-1)
m = 1/1
m = 1
The y intercept is b = -3. You can see this if you extend the line until it crosses the y axis.
Alternatively, plug in (x,y) = (1,-2) and m = 1 into y = mx+b to find that b = -3
So y = mx+b turns into y = 1x+(-3) or just y = x-3
We combine both parts to end up with
This is only graphed when (note the closed or filled in circle for the endpoint of this portion).
===================================================
Problem 5
Answer:
<h3>------------------
Work Shown:
y = |x| .... parent function
y = |x+3| ... shift 3 units to the left
y = (1/2)*|x+3| .... vertically compress by factor of 1/2
f(x) = (1/2)*|x+3|
------
Break that down into a piecewise function
when x < -3, then y = -(1/2)(x+3)
when , then y = (1/2)(x+3)
I'm using the rule that y = |x| turns into y = -x when x < 0 and y = x when
So that is how we get as the piecewise function.