The first set:
3x + 2y = 2 ---1)
5x + 4y = 6 ---2)
From 1), multiply all by 2, 6x + 4y = 4 ---3)
3) - 2),
6x + 4y - (5x + 4y) = 6 - 4
6x + 4y - 5x - 4y = 2
x = 2
Sub in x = 2 into 1),
3(2) + 2y = 2
2y = -4
y = -2
(2 , -2)
The second set:
3x + 2y = 2 ---1)
11x + 8y = 10 ---2)
From 1), multiply all by 4, 12x + 8y = 8 ---3)
3) - 2),
12x + 8y - (11x + 8y) = 8 - 10
12x + 8y - 11x - 8y = -2
x = -2
From this x value alone, we can tell that these two linear systems do NOT have the same solution as they meet at different coordinates.
Hope this helped! Ask me if there's any working from here that you don't understand! :)
Answer:

Step-by-step explanation:

If it’s only reflected over the y axis then only the x values change
Answer:
See Below.
Step-by-step explanation:
We are given that ΔAPB and ΔAQC are equilateral triangles.
And we want to prove that PC = BQ.
Since ΔAPB and ΔAQC are equilateral triangles, this means that:

Likewise:

Since they all measure 60°.
Note that ∠PAC is the addition of the angles ∠PAB and ∠BAC. So:

Likewise:

Since ∠QAC ≅ ∠PAB:

And by substitution:

Thus:

Then by SAS Congruence:

And by CPCTC:

Answer:
B. A teacher compares the pre-test and post-test scores of students
Step-by-step explanation:
the answer is true, because it is a good example to compare the tests between students, we know that a matching pair design is a random model and is used when the experiment allows grouping subjects in pairs based on a variable and each pair will receive randomly a different handling, the answer A is not true because in the example all students are uniformly averaged and the variable is not correlated with subgroups, option c is incorrect because the variable was not randomized and generates classification bias and the option d is incorrect because the teacher compares a small sample as her class with a score of a total sample, but does not intervene on her students when selecting the corresponding group