Solve the system of equations by graphing. Check your solution. 5x+y=14 2x-y=7
1 answer:
ANSWER
(3,-1)
EXPLANATION
The given system of equations is :
5x+y=14
2x-y=7
For the first equation when x=0,
5(0)+y=14
y=14
The y-intercept is (0,14).
When y=0, 5x+0=14
The x-intercept is (2.8,0).
We plot the intercepts and draw a straight line through them to obtain the graph of 5x+y=14 as shown in the attachment.
We repeat the same process for the second equation 2x-y=7.
The intercepts are (0,-7) and (3.5,0)
The solution to the system is the point of intersection of the two straight lines.
The two straight lines intersected at
(3,-1)
Hence x=2, y=-1
(3,-1)
EXPLANATION
The given system of equations is :
5x+y=14
2x-y=7
For the first equation when x=0,
5(0)+y=14
y=14
The y-intercept is (0,14).
When y=0, 5x+0=14
The x-intercept is (2.8,0).
We plot the intercepts and draw a straight line through them to obtain the graph of 5x+y=14 as shown in the attachment.
We repeat the same process for the second equation 2x-y=7.
The intercepts are (0,-7) and (3.5,0)
The solution to the system is the point of intersection of the two straight lines.
The two straight lines intersected at
(3,-1)
Hence x=2, y=-1
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Answer:
Y = 3x^x is a graph that has exponential growth while y = 3^-x has exponential decay.
Y = 3x^x (-∞, 0) and (∞, ∞).
Y = 3x^-x (-∞, ∞) and (∞, 0).
Step-by-step explanation:
The infinity symbols were being used to represent the x and y values of each graph. I will call y = 3^x "graph 1" and y = 3^-x "graph 2".
When graph 1 had positive ∞ for its x value, its y value was reaching towards positive ∞. When its x was reaching for negative ∞, its y was going for 0.
For graph 2, however, when its x was reaching for positive ∞, its x was reaching for 0. When its x was reaching for negative ∞, its y was going for positive ∞.
Here's an image of the graphs:
