One positive number is 6 more than twice another. If their product is 1856, find the numbers.

Question

One positive number is 6 more than twice another. If their product is 1856, find the numbers.

Answer 1

Create two equations from the information given, then substitute one into the other and solve.

 

Info #1

 

One positive number (call it "y') is 6 more than twice another (call it "x").  "6 more" means add 6.  "Twice" means multiply by 2.  Therefore...

 

  Equation #1:  y = 2x + 6

 

Info #2

 

Their product is 1856.  "Product" means multiply.  Therefore...

 

  Equation #2:  x _* y = 1856

 

Substitute

 

Substitute "2x + 6" from equation #1 (since that is what y equals) for the y in equation #2 to get an equation all in terms of x, so you can solve for it.

 

  New Equation:  x(2x + 6) = 1856

 

Solve

 

  x (2x + 6) = 1856       Perform the multiplication by distributing the x

 

  2x² + 6x = 1856         Divide entire equation by 2, the coefficient of x²

 

  x² + 3x = 928             Set equation (it is a quadratic) to 0 in order to solve by subtracting 928 from both sides

 

  x² + 3x - 928 = 0        We can solve this using the Quadratic Formula or by Factoring.

 

  Factoring:

 

    x² + 3x - 928 = 0      Solve by Factoring

 

    Since the 928 is preceded by a minus sign (-), we know one factor must be "x +" and the other "x -", so factor the equation as follows...

 

    (x + n₁) (x - n₂) = 0

 

    We also know that the product of n_{1 *} n₂ must = -928 and their sum must = +3 (the coefficient of the x term).

    So what 2 numbers could multiply to be -928 but add to be +3 ??

    For a product to be negative means one factor has to be positive and the other factor negative.

    Then for a positive factor and a negative factor to sum to +3 means their absolute values have to be 3 apart with the positive factor being the larger.  Hmmm...

    I know 30 _* 30 = 900 (a little low and not 3 apart) so how about 31 and 34?  But 31 _* 34 = 1054, too big.

    So let's go the other way... 29 and 32.  29 _* 32 = 928.  Bingo!

    But remember their product has to be negative while their sum is positive, so make 32 (the larger one) positive and 29 negative.

    Since n₁ is positive and n₂ is negative in our factoring above, we now have...

 

    (x + 32) (x - 29) = 0

 

    To check our factoring, let's multiply it out.  We get...

 

    x² - 29x + 32x - 928 = 0 that simplifies to...

 

    x² + 3x - 928 = 0

 

    Now (finally) solve for x.

 

    If (x + 32) (x - 29) = 0 then either (x + 32) = 0 or (x - 29) = 0

 

    x + 32 = 0 solves to x = -32 but -32 is not a positive number (requirement) so the solution must be the other.

 

    x - 29 = 0 solves to x = 29.  That's one of our numbers.  What's the other?  Go back to Equation #1 or #2.

 

    y = 2x + 6, set x = 29                                       x _* y = 1856, set x = 29

 

    y = 2 (29) + 6 = 58 + 6 = 64   That's our y!      29y = 1856, so y = 1856/29 = 64   That's our y!

 

    So x = 29 and y = 64 solves the problem!

 

  Quadratic Formula:

 

    x² + 3x - 928 = 0      Solve using the Quadratic Formula

 

    Formula

 

    For ax² + bx + c = 0, we solve for x as

 

    x = [-b ± √ (b² - 4ac)] / 2a

 

    So let's solve.  Using our equation above, a = 1, b = 3, and c = -928.  (from x² +3x - 928 = 0)

 

    x = [-3 ± √ (3² - 4(1)(-928))] / 2(1)  =  [-3 ± √ (9 + 3712)] / 2  =  [-3 ± √3721] / 2  =  [-3 ± 61] / 2

 

    Therefore x = (-3 + 61) / 2  or  (-3 - 61) / 2  so...

    x = 58/2 = 29 or x = -64/2 = -32, but again x has to be positive, so x = 29!

 

    Substitute x = 29 into either Equation #1 or #2 (just like with Factoring) and we get y = 64!

 

    y = 2x + 6 = 2(29) + 6 = 58 + 6 = 64   or    x _* y = 1856 → 29y = 1856 → y = 1856/29 = 64!

 

Done!  Solved both by Factoring and the Quadratic Formula.