There's some unknown (but derivable) system of equations being modeled by the two lines in the given graph. (But we don't care what equations make up these lines.)
There's no solution to this particular system because the two lines are parallel.
How do we know they're parallel? Parallel lines have the same slope, and we can easily calculate the slope of these lines.
The line on the left passes through the points (-1, 0) and (0, -2), so it has slope
(-2 - 0)/(0 - (-1)) = -2/1 = -2
The line on the right passes through (0, 2) and (1, 0), so its slope is
(0 - 2)/(1 - 0) = -2/1 = -2
The slopes are equal, so the lines are parallel.
Why does this mean there is no solution? Graphically, a solution to the system is represented by an intersection of the lines. Parallel lines never intersect, so there is no solution.
We will solve this by suing simultaneous equations,
⇒ 5s + 3j = 87
4s + 2j = 64
Multiply the first equation with 4 and the second one with 5, this is to get one of the values equal so that we can cancel them out,
⇒ (5s + 3j = 87) × 4
(4s + 2j = 64) × 5
∴ ⇒ 20s + 12j = 348
20s + 10j = 320
Subtract both the equations. This is how your result (after subtraction) should look like,
⇒ 2j = 28
∴ ⇒ j = $14
Now replace the value of 'j' in one of the original equations,
⇒ 4s + 2(14) = 64
⇒ 4s + 28 = 64
⇒ 4s = 64 - 28
⇒ 4s = 36
∴ ⇒ s = $9
Therefore, one pair of jeans cost $14 and a shirt costs $9
Hope you understood! Feel free to ask me if you didn't understand a step.
Answer:
p^3 / q^12
Step-by-step explanation:
p^6 q^4
------------------
p^3 q^16
We know a^b / a^c = a^(b-c)
First with variable p
p^6 / p ^3 = p^(6-3) = p^3
Then with variable q
q^4 / q^16 = q^(4-16) = q^-12 and a^-b = 1/ a^b = 1 /q^12
p^3 * 1/ q^12
p^3 / q^12
The student had a 40% error estimate
