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Nana76 [90]
4 years ago
7

FIND THE SOLUTION OF EACH EQUATION IF THE REPLACEMENT SETS ARE

Mathematics
1 answer:
Yuliya22 [10]4 years ago
3 0
<span>1) There are no solutions for this equation in the solution set.

2) The solution set is {5} because 15/5 = 3

3) The solution set is {8} because 31 = 3 (8) + 7 \\ 31 = 24 + 7 \\ 31 = 31

4) The solution set is {20} because 20 - 13 = 7

5) The solution set is {66} because 22 x 3 = 66

6) A = 32 - 9(2) \\ A = 32 - 18\\ A = 14

7) 12 \times \frac{5}{15} - 3 = Y \\\\ 12 \times \frac{1}{3} -3 = Y \\\\ 4 -3 = Y \\ Y =1

8) </span>27 + \frac{5}{16} = G\\\\G = \frac{432 + 5}{16}\\\\G = \frac{437}{16}\\\\G = 27 \frac{5}{16}

9) R = \frac{9(6)}{ ( 8 + 1 ) 3}\\\\R = \frac{54}{9(3)} \\\\R = \frac{54}{27}\\\\ R = 2

10) 15,579 + 6,220 + 18,995 = G\\\\ G = 40,794
<span>
11) The solution set is {6, 7} because:
 6 -2 < 6, which is 4 < 6
AND
7 - 2 < 6, which is 5 < 6

12) </span><span>The solution set is {9, 11} because:
</span>3 ≥ 25/9, which is 3 ≥ <span>2.77777777778
</span>AND
3 ≥ 25/11, which is 3 ≥ 2.2727272727

Hope this helps :D
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