How can I find the solution of the following system of equations?

Question

1. m - n = 8
2m + n = 1

2. a + 2b = 9
4a + 3b = 2

3. p - q = 3
3p + 2q = 4

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Please solve them by comparison or substitution.

Answer 1

1. We know m - n = 8 and 2m + n = 1

m - n = 8
  +n  +n get one variable alone
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m = 8 + n now we also know m=8+n

So substitute
2 * m + n = 1
2 * (8+n) +n = 1
16 + 2n + n = 1
16 + 3n = 1
  -16     -16
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    3n = -15
    /3    /3 Divide both sides by 3
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    n = -5

So we know n = - 5 and we still know m - n = 8
so m - (-5) = 8
  m + 5 = 8
   m = 3

So m = 3 and n = -5

Do the rest yourself

Answer 2

m-n=8 ……….(1) 2m+n=1………….(2)
adding (1) and (2) 3m=9 and so m=3
substituting m=3 in(1)3-n=8 and so n=-5
soothe solution set is {3,-5}
check3-(-5)=8 and 2(3)-5=1

Answer 3

1. m-n = 8 (equation 1)
2m+n = 1(equation 2)

solve for n in terms of m in equation 1.
-n=-m+8
n=m-8 (equation 3)

substitute m-8 for n in equation 2.
2m + (m-8) =1
3m-8=1
3m=1+8
3m=9
m=3

substitute 3 for m in equation 3.
n=m-8
n=3-8
m=-5

solution: ( 3,-5)
check:
m-n=8
3-(-5)=8
3+5=8
8=8


i can't answer#s 2 & 3...sorry!!!

Answer 4

m=3
m= negative 5

p=2
q= negative 1