After re-writing 2 equations into y-intercept form, the equations are the same (with the same slope amd y -intercept.) How many possible solutions can there be for these systems of equations?
A.one
B.infinie
C.zero
A. one
Write y+4x=2 in slope-intercept form.
A.y= - 4x+2
B.y=4x+2
C.y=4x-2
To write the equation \(y + 4x = 2\) in slope-intercept form, we need to solve for \(y\).
Starting with \(y + 4x = 2\), we can isolate \(y\) by subtracting \(4x\) from both sides:
\(y = -4x + 2\)
Therefore, the equation in slope-intercept form is:
A. \(y = -4x + 2\)
y=4x-6
6x+2y=2
If I rewrite the second equation as 6x+2(4x-6)=2, what method am I using?
A.eliminating y
B.substituting y for: 4x-6
C.adding 4x-6 to y
If you rewrite the second equation as \(6x+2(4x-6)=2\), you are using method:
B. substituting \(y\) for \(4x-6\)
-10x-3y--58
5x + 3y=38
Why should I add these equations?
A.For fun!
B.To eliminate both the variables
C.to eliminate y and solve for x
In the given equations:
\[-10x - 3y = -58\]
\[5x + 3y=38\]
Adding these equations will eliminate the variable \(y\) which makes it easier to solve for \(x\). Therefore, the correct answer is:
B. To eliminate both variables
Which equation represents the real-world situation below. What do your variables represent?
A.250 m +15=20 (m=weeding)
B.m+w=14 (m=mowing and w= weeding)
C.250w =14 (w=mowing)
D.Other:
The real-world situation you provided involves both mowing and weeding, so the correct equation representing this situation is:
B. \(m + w = 14\), where \(m\) represents mowing and \(w\) represents weeding.