SOLUTION: I have done this problem over and over and I can't figure it out.
The sum of three times a first number and twice a second number is 43. If the second number is subtracted from tw
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Question 101789This question is from textbook
: I have done this problem over and over and I can't figure it out.
The sum of three times a first number and twice a second number is 43. If the second number is subtracted from twice the first number, the result is -4. Find the numbers. Can you please help me.
This question is from textbook
Answer by Fombitz(32388) (Show Source): You can put this solution on YOUR website!
Let’s call the first number A, the second number B.
(1) 3A + 2B = 43
(2) 2A – B = -4
Let’s use equation (2) to solve for B in terms of A and then we’ll substitute in equation (1) to solve for A.
(2) 2A – B = -4
2A-B+B=-4+B Use the additive inverse of (-B) or B
2A=-4+B Simplify
2A+4=-4+4+B Use the additive inverse of (-4) or 4
2A+4=B or
(3) B=2A+4 We’ll call that equation (3), that’s B in terms of A. Really it’s equation (2) re-arranged.
Now let’s use this answer in equation (1)
(1) 3A + 2B = 43
3A + 2(2A+4) = 43 Plug in your value for B.
3A + 2(2A) + 2(4) = 43 Use the Distributive property
3A + 4A + 8 = 43 Simplify
7A +8 = 43
7A + 8 – 8 = 43 – 8 Use the additive inverse of (8) or -8
7A = 35
7A/7=35/7 Use the multiplicative inverse of (7) or 1/7
A=5 Use your answer for A in equation (1),(2), or (3) and solve for B.
Let’s use equation(3), it’s most direct.
(3) B=2A+4
B=2(5)+4
B=14
The answers are A=5, B=14.
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