Asked by tt
The sum of Sharon's and John's ages is 70
. Sharon is 4
times as old as John.
If you let s=
Sharon's age and j=
John's age, then the problem can be represented by a system of equations. Which of the following shows a graph of this system and the solution to this problem?
A graph shows S on the x-axis, J on the y-axis, and two lines that intersect. A line with a positive slope touches the y-axis at (0, 70) and continues to (just past 30, 100). A line with a positive slope touches the origin at (0, 0) and continues to (between 20 and 30, 100). The lines intersect at (just past 20, just past 90).
A graph shows S on the x-axis, J on the y-axis, and two lines that intersect. A line with a negative slope touches the y-axis at (0, 70) and the x-axis at (70, 0). A line with a positive slope touches the origin at (0, 0) and continues to (100, just above 20). The lines intersect at (between 50 and 60, between 10 and 20).
A graph shows S on the x-axis, J on the y-axis, and two lines that intersect. A line with a positive slope touches the x-axis at (70, 0) and continues to (100, just before 30). A line with a positive slope touches the origin at (0, 0) and continues to (100, between 20 and 30). The lines intersect at (just past 90, just past 20).
A graph shows S on the x-axis, J on the y-axis, and two lines that intersect. A line with a negative slope touches the y-axis at (0, 70) and touches the x-axis at (70, 0). A line with a positive slope touches the origin at (0, 0) and continues to (between 20 and 30, 100). The lines intersect at (between 10 and 20, between 50 and 60).
All Answers 1
Answered by
GPT-5 mini
AI
Set up the system:
s + j = 70 (so j = 70 − s, a line with slope −1, y‑intercept 70 and x‑intercept 70)
s = 4j (so j = s/4, a line through the origin with positive slope 1/4)
Solve: substitute j = s/4 into s + j = 70 ⇒ s + s/4 = 70 ⇒ (5/4)s = 70 ⇒ s = 56, j = 14.
Choice 2 matches (negative‑slope line through (0,70) and (70,0), positive‑slope line through the origin, intersection between 50 and 60 for s and between 10 and 20 for j).
s + j = 70 (so j = 70 − s, a line with slope −1, y‑intercept 70 and x‑intercept 70)
s = 4j (so j = s/4, a line through the origin with positive slope 1/4)
Solve: substitute j = s/4 into s + j = 70 ⇒ s + s/4 = 70 ⇒ (5/4)s = 70 ⇒ s = 56, j = 14.
Choice 2 matches (negative‑slope line through (0,70) and (70,0), positive‑slope line through the origin, intersection between 50 and 60 for s and between 10 and 20 for j).
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