Can anyone please help me with these 2 college algebra questions

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There is a number of ways to do this.
Here's one.

➊ (³∕₂)x − (⅓)y = -18            ← Notice that multiplying by 2 gets a 3x term
➋ (¾)x + (²∕₉)y = -9            ← Notice that multiplying by 4 gets a 3x term

➊ 3x − (²∕₃)y = -36            ← Multiplied both sides by 2
➋ 3x + (⁸∕₉)y = -36            ← Multiplied both sides by 4

➊                      3x = (²∕₃)y - 36            ← Now, substitute this into ➋ for 3x
➋  (²∕₃)y - 36 + (⁸∕₉)y = -36
            (²∕₃)y + (⁸∕₉)y = 0           ← Notice there's no need to add ²∕₃ and ⁸∕₉
                            y = 0  ✔       ← Now, substitute y=0 into the original ➊ or ➋


➊ (³∕₂)x − (⅓)y = -18
   (³∕₂)x − (⅓)·0 = -18
              (³∕₂)x = -18
                   x = -18(⅔)
                   x = -12  ✔


                ANSWER
                  x = -12
                  y = 0


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➌ 0.1x + 0.002y = 0.04
➍               5x+y = 2  ⇒  y = 2-5x           ← Now, substitute this into ➌


➌ 0.1x + 0.002(2-5x) = 0.04           ← Substituted (2-5x) in for y
          100x + 2(2-5x) = 40           ← Multiplied both sides by 1,000
          100x + 4 - 10x = 40
                           90x = 36
                               x = 36/90
                               x = ⅖ or 0.4  ✔           ← Substitute this into ➍ to get y


➍               5x+y = 2
              5(⅖)+y = 2
                  2 + y = 2
                        y = 0  ✔


               ANSWER
                  x = 0.4
                  y = 0


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