Stephen H. answered • 11/26/14
AP Calculus AB/BC, GED & more - M.Ed. w/ 10+ yrs of experience
Hi Kyrsten, Since (3x-5)/2=y and y=(2x+4)/5 we can substitute, or replace, the first y with (2x+4)/5. This gives us (3x-5)/2=(2x+4)/5. Phew, ok. Now we're just dealing with x's and constants (no y's now!). That's better. From here, we have two fractions equaling each other. We can solve problems in this form by "cross multiplying." This technique allows us to get rid of fractions. That means we'll multiply (3x-5) by the 5 on the other side, and (2x+4) by the 2 on the other side. This gives (3x-5)*5=(2x+4)*2. Distribute the 5 and you'll get 15x-25. Distribute the 2 on the right side and you'll get 4x+8. Together, this gives 15x-25=4x+8. Subtract 4x from both sides. Now, 11x-25=8. Add 25 to both sides. Now, 11x=33. Divide both sides by 11. Now, x=3. We can substitute (or replace) 3 for x in either of the first two equations. Let's take the first equation (3x-5)/2=y. Replace the x with a 3. (3*3-5)/2=y, which is (9-5)/2=y, which is 4/2=y, or 2=y. To write your answer in (x, y) form, since x=3 and y=2, we'd have (3, 2).
