Two-variable word problems: Systems of Equations. Please Help!!!?

let x = first number and y = second number

x = 16 + y

we know that y is the smaller number.

(2/3)x - y = 1/4 (x + y)

just sub x = 16 + y

(2/3)(16 + y) - y = 1/4 (16 + y + y)

multiply both sides by 12 (common factor or whatever it's called)

8(16 + y) - 12y = 3(16 + 2y) ?(if you're wondering where i got the 8, it's 2/3 * 12 = 8, and i got the 3 from 1/4 * 12 = 3)

128 + 8y - 12y = 48 + 6y

128 - 4y = 48 + 6y

80 = 10y

y = 8

x = 16 + 8 = 24

This is just too many disorders. You have got to uncover out find out how to do one, and get the leisure your self. Unless any one is inclined to do all of it. a million. A + B = ninety A - B = 18 Two equations, 2 unknowns. A approach of equations. All those paintings the equal approach. There are a few methods to clear up a approach. You can upload one equation to the opposite - upload like phrases. In this example you'll be able to see that B + (-B) goes to cancel, that is quality given that you're going to emerge as with simplest A. You get: 2A = ninety + 18 2A = 108 A = fifty four Plug into both equation to uncover B. fifty four - B = 18, so B = fifty four -18 = 36. The different solution to clear up is specific one quantity in phrases of the opposite in a single equation, and replacement into the opposite. Using the primary equation, A = ninety - B. Plug into the moment: (ninety - B) - B = 18 ninety - 2B = 18 2B = ninety - 18 = seventy two B = 36 Now replacement in both equation to uncover that A = fifty four