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vlabodo [156]
4 years ago
5

Find three consecutive integers such that when three times the third is increased by the first, the result is 8 less than five t

imes the second.
Mathematics
1 answer:
nexus9112 [7]4 years ago
8 0

Answer:

<h3>            9, 10 and 11</h3>

Step-by-step explanation:

x - the smallest of three consecutive integers (the first)

x+1 - the next of three consecutive integers (the second)

x+2 - the last of three consecutive integers (the third)

3(x+2) - three times the third integer

<u>3(x+2)+x </u> - three times the third increased by the first

5(x+1)   - five times the second tnteger

<u>5(x+1)-8 </u>  - 8 less than five times the second

3(x + 2) + x = 5(x + 1) - 8

3x + 6 + x = 5x + 5 - 8

   4x + 6 = 5x - 3

       -4x        -4x

        6 = x - 3

         x = 9

x+1 = 10

x+2 = 11

Check:  3×11+9 = 33+9 = <u>42</u>,   5×10-8=50-8 = <u>42</u>

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