Here are the solutions for each question in sequence:
Question 1:
Solve this system of equations:
\[
−2x + y = 1 \quad \text{(1)} \
−4x + y = −1 \quad \text{(2)}
\]
Subtract equation (1) from equation (2):
\[
(-4x + y) - (-2x + y) = -1 - 1 \implies -2x = -2 \implies x = 1
\]
Substituting \(x = 1\) into equation (1):
\[
−2(1) + y = 1 \implies -2 + y = 1 \implies y = 3
\]
So, the solution is \((1, 3)\).
Question 2:
From the equation \(−x − 2y = −10\), to isolate \(x\):
\[
-x = -10 + 2y \implies x = 10 - 2y
\]
Therefore, the expression to substitute into the first equation \(3x + y = 8\) is \(10 - 2y\).
So, the correct answer is:
−2y + 10
Question 3:
Using the equations:
\[
x = 3y \quad \text{(1)} \
x = 10 + 2y \quad \text{(2)}
\]
From equation (1), substitute into equation (2):
\[
3y = 10 + 2y \implies 3y - 2y = 10 \implies y = 10 \
x = 3(10) = 30
\]
So, Sam is \(30\) years old and Harry is \(10\) years old.
The correct option is:
Sam is 30 years old, and Harry is 10 years old.
Question 4:
We have:
\[
y = -2x + 1 \quad \text{(1)} \
4x + 2y = -1 \quad \text{(2)}
\]
Substituting expression for \(y\) from equation (1) into (2):
\[
4x + 2(-2x + 1) = -1 \implies 4x - 4x + 2 = -1 \implies 2 = -1 \quad \text{(Contradiction)}
\]
Thus, this system has No solution.
Question 5:
Let \(n\) be the number of notebooks and \(t\) be the number of T-shirts:
\[
n + t = 90 \quad \text{(1)} \
6n + 12t = 600 \quad \text{(2)}
\]
From equation (1), \(t = 90 - n\). Substituting into (2):
\[
6n + 12(90 - n) = 600 \
6n + 1080 - 12n = 600 \
-6n + 1080 = 600 \
-6n = -480 \
n = 80
\]
So, the answer is:
80
Question 6:
Let \(x\) be the number of children and \(y\) be the number of adults:
\[
x + y = 320 \quad \text{(1)} \
7x + 10y = 2900 \quad \text{(2)}
\]
The correct system of equations modeling the scenario is:
\[
\text{Correct answer: } 7x + 10y = 2900 \quad \text{and } x + y = 320
\]
Question 7:
Given the equations:
\[
x = 60 + 4y \quad \text{(1)} \
7x + 12y = 500 \quad \text{(2)}
\]
Substituting (1) into (2):
\[
7(60 + 4y) + 12y = 500 \
420 + 28y + 12y = 500 \
40y = 500 - 420 \
40y = 80 \
y = 2 \
x = 60 + 4(2) = 68
\]
Thus, the answer is:
(68, 2)
Question 8:
From the system:
\[
4x = 8 + 3y \
x - 2y = -3
\]
First equation can be rearranged as:
\[
4x - 3y = 8
\]
Rearranging the second gives:
\[
x = 2y - 3
\]
For substitution:
Substituting \(x\) into \(4x - 3y = 8\):
\[
4(2y - 3) - 3y = 8 \
8y - 12 - 3y = 8 \
5y - 12 = 8 \
5y = 20 \
y = 4 \quad x = 2(4) - 3 = 5
\]
Thus, temperatures are:
City A was 5° C, and City B was 4° C.
Question 9:
Let \(p\) be the price of a bag of popcorn and \(c\) be the price of a cup.
Using the equations:
\[
4p + 5c = 50 \
6p + 4c = 54
\]
Multiplying the first equation by 4 and the second by 5 gives:
\[
16p + 20c = 200 \
30p + 20c = 270
\]
Subtract the first from the second:
\[
14p = 70 \
p = 5
\]
So, the price of one bag of popcorn is:
$5
Question 10:
Using the system:
\[
−2x + 4y = 6 \quad \text{(1)} \
y = 2x + 3 \quad \text{(2)}
\]
Substituting (2) into (1):
\[
−2x + 4(2x + 3) = 6 \
−2x + 8x + 12 = 6 \
6x = 6 - 12 \
6x = -6 \
x = -1 \quad y = 2(-1)+3 = 1
\]
Final solution:
(−1, 1)
9 answers