find three consecutive even integer such that the sum of twice the first integer four more than the second and seven less that the third integer is 27 find the integers.
Answers
Answer:
Required integers are 10, 16 & 20.
Explanation:
Given that:
Sum of twice the first integer four more than the second integer and seven less that the third integer is 27.
To Find:
What are the three integers?
Solution:
Let the first integer be x and second integer be y.
According to the question,
Sum of twice the first integer is 7 less than third integer which is 27.
Therefore,
⇒ \sf 2x = 27 - 72x=27−7
⇒ \sf 2x = 202x=20
⇒ \sf x = {\cancel{\dfrac{20}{2}}}x=
2
20
➠ \bf\red{x = 10}x=10
∴ First integer (x) is \boxed{\bf{10}}
10
Also,
Sum of twice the first integer is 4 more than the second integer.
Therefore,
⇒ \sf 2x = y + 42x=y+4
Put x = 10 in above equation we get,
⇒ \sf 2\:\times\:10 = y + 42×10=y+4
⇒ \sf 20 = y + 420=y+4
⇒ \sf y = 20 - 4y=20−4
➠ \bf\green{y = 16}y=16
∴ Second integer (y) is \boxed{\bf{16}}
16
Hence,
\sf \:\red{\bigstar}\:Required\:integers\:\leadsto\:{\large{\boxed{\tt{\purple{10,\:16,\:\&\:27}}}}}★Requiredintegers⇝
10,16,&27
Answer:
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