Question

unit: systems of equations and inequalities
progress:
question id: 108564
the movement of the progress bar may be uneven because questions can be worth more or less (including zero) depending on your answer.
john is 70 years younger than sharon. sharon is 4 times as old as john.
if you let ( s = ) sharons age and ( j = ) johns 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?
four graphs are shown here (omitted as per instruction)

Explanation:

Step1: Formulate the equations

From "John is 70 years younger than Sharon", we get \( s = j + 70 \) (or \( j=s - 70\)). From "Sharon is 4 times as old as John", we get \( s = 4j \).

Step2: Solve the system of equations

Substitute \( s = 4j \) into \( s=j + 70 \), we have \( 4j=j + 70 \). Subtract \( j \) from both sides: \( 4j-j=j + 70-j \), so \( 3j = 70\)? Wait, no, wait: Wait, "John is 70 years younger than Sharon" means \( j=s - 70 \), and "Sharon is 4 times as old as John" means \( s = 4j \). Substitute \( s = 4j \) into \( j=s - 70 \), we get \( j = 4j-70 \). Subtract \( 4j \) from both sides: \( j-4j=4j - 70-4j \), so \( - 3j=-70 \), then \( j=\frac{70}{3}\approx23.33 \)? Wait, no, maybe I mixed up the variables. Wait, the problem says "John is 70 years younger than Sharon", so Sharon's age \( s=j + 70 \). And "Sharon is 4 times as old as John", so \( s = 4j \). So set \( 4j=j + 70 \), then \( 4j-j=70 \), \( 3j = 70 \), \( j=\frac{70}{3}\approx23.33 \), \( s = 4\times\frac{70}{3}=\frac{280}{3}\approx93.33 \). But looking at the graphs, the fourth graph (the one with the two lines, one decreasing? Wait no, wait the first equation \( s=j + 70 \) in terms of \( j \) (if \( j \) is y - axis and \( s \) is x - axis) would be \( j=s - 70 \), a line with slope 1, y - intercept - 70? No, maybe the axes are \( j \) (John's age) on y - axis and \( s \) (Sharon's age) on x - axis. Wait, the equations:

Equation 1: \( s=j + 70 \) (Sharon's age = John's age + 70)

Equation 2: \( s = 4j \) (Sharon's age = 4*John's age)

To graph these, for Equation 1: when \( j = 0 \), \( s=70 \); when \( s = 0 \), \( j=-70 \) (not practical for age, but the line has slope 1). For Equation 2: when \( j = 0 \), \( s = 0 \); when \( j = 10 \), \( s = 40 \), slope 4.

The solution is when \( 4j=j + 70\Rightarrow3j = 70\Rightarrow j=\frac{70}{3}\approx23.33 \), \( s=\frac{280}{3}\approx93.33 \). Now looking at the graphs, the fourth graph (the one with the two lines, one with positive slope 1 (from \( s=j + 70 \)) and one with positive slope 4 (from \( s = 4j \)) intersecting at \( j\approx23.33 \), \( s\approx93.33 \). Wait, but maybe I made a mistake in variable assignment. Wait, the problem says "let \( s=\) Sharon's age and \( j=\) John's age". So the first equation: \( j=s - 70 \) (John is 70 younger than Sharon), and the second equation: \( s = 4j \). So substituting \( s = 4j \) into \( j=s - 70 \), we get \( j = 4j-70\Rightarrow - 3j=-70\Rightarrow j=\frac{70}{3}\approx23.33 \), \( s = 4\times\frac{70}{3}=\frac{280}{3}\approx93.33 \).

Now, looking at the graphs:

- The first graph: one line increasing slowly (slope 1/4 maybe) and one decreasing (slope - 1). No.

- The second graph: two lines with positive slopes, one steeper (slope 4) and one less steep (slope 1), intersecting at \( j\approx23 \), \( s\approx93 \). That matches. Wait, no, the second graph's axes: if \( j \) is y - axis and \( s \) is x - axis, Equation 1: \( j=s - 70 \) (slope 1, y - intercept - 70) and Equation 2: \( j=\frac{s}{4} \) (slope 1/4, y - intercept 0). Wait, maybe I had the axes reversed. If \( j \) is on the y - axis and \( s \) is on the x - axis, then:

Equation 1: \( j=s - 70 \) (line with slope 1, x - intercept 70, y - intercept - 70)

Equation 2: \( j=\frac{s}{4} \) (line with slope 1/4, passing through origin)

The intersection is at \( s - 70=\frac{s}{4}\Rightarrow4s-280 = s\Rightarrow3s = 280\Rightarrow s=\frac{280}{3}\approx93.33 \), \( j=\frac{70}{3}\approx23.33 \). Now, looking at the fourth graph (the one with the two lines, one with slope 1 (from \( j=s - 70 \)) and one with slope 1/4 (from \( j=\frac{s}{4} \)) intersecting at \( s\approx93 \), \( j\approx23 \)). Wait, the fourth graph has a line that looks like \( j=s - 70 \) (going from (70,0) to (0, - 70) but shifted? No, maybe the axes are \( j \) (y) and \( s \) (x), and the two lines: one is \( s = j + 70 \) (so \( j=s - 70 \)) and one is \( s = 4j \) (so \( j=\frac{s}{4} \)). The intersection point is where \( s - 70=\frac{s}{4}\), which we solved as \( s=\frac{280}{3}\approx93.33 \), \( j=\frac{70}{3}\approx23.33 \). The fourth graph (the last one) has the two lines: one with positive slope 1 (from \( s=j + 70 \)) and one with positive slope 4 (from \( s = 4j \))? Wait, no, if \( s \) is x - axis and \( j \) is y - axis, \( s = 4j \) is \( j=\frac{s}{4} \) (slope 1/4), and \( s=j + 70 \) is \( j=s - 70 \) (slope 1). So the line \( j=s - 70 \) has a steeper slope than \( j=\frac{s}{4} \). Wait, no, slope of \( j=s - 70 \) is 1, slope of \( j=\frac{s}{4} \) is 1/4. So the line \( j=s - 70 \) is steeper. Now, looking at the fourth graph, there is a line with slope 1 (steeper) and a line with slope 4? No, maybe I messed up the equations. Wait, let's re - express the equations correctly.

Let me start over:

Let \( s=\) Sharon's age, \( j=\) John's age.

1. John is 70 years younger than Sharon: \( j=s - 70 \) (John's age = Sharon's age - 70)

2. Sharon is 4 times as old as John: \( s = 4j \) (Sharon's age = 4*John's age)

Now, substitute \( s = 4j \) into \( j=s - 70 \):

\( j=4j - 70 \)

Subtract \( 4j \) from both sides:

\( j-4j=4j - 70-4j \)

\( - 3j=-70 \)

Divide both sides by - 3:

\( j=\frac{70}{3}\approx23.33 \)

Then \( s = 4j=\frac{280}{3}\approx93.33 \)

Now, to graph the equations:

- Equation 1: \( j=s - 70 \). This is a linear equation in slope - intercept form (\( j=1\times s-70 \)) with slope \( m = 1 \) and y - intercept (when \( s = 0 \)) \( j=-70 \) (not relevant for age, but mathematically, the line goes through (70,0) when \( j = 0 \), since \( 0=s - 70\Rightarrow s = 70 \))

- Equation 2: \( j=\frac{s}{4} \). This is a linear equation with slope \( m=\frac{1}{4} \) and y - intercept \( 0 \), passing through the origin (0,0)

The point of intersection is at \( s=\frac{280}{3}\approx93.33 \), \( j=\frac{70}{3}\approx23.33 \)

Now, looking at the four graphs:

- The fourth graph (the one with the two lines, one with slope 1 (from \( j=s - 70 \)) and one with slope \( \frac{1}{4} \) (from \( j=\frac{s}{4} \)) intersecting at \( s\approx93 \), \( j\approx23 \)) is the correct one. Because the line \( j=s - 70 \) will have a steeper slope (slope 1) than \( j=\frac{s}{4} \) (slope \( \frac{1}{4} \)), and they intersect at the solution point we found.

Answer:

The fourth graph (the one with the two lines, one with a steeper positive slope (representing \( s = j + 70 \)) and one with a less steep positive slope (representing \( s = 4j \)) intersecting at \( j\approx23.33 \) and \( s\approx93.33 \)) is the correct graph and solution. (Assuming the fourth option in the multiple - choice is the one with these characteristics, e.g., if the options are labeled as A, B, C, D, and D is the fourth graph, then the answer is D. The solution to the problem is John's age \( j=\frac{70}{3}\approx23.33 \) years and Sharon's age \( s=\frac{280}{3}\approx93.33 \) years.)