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
Algebra ->
Equations
-> 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
Log On
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
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.
