Question 1178145
x = number of tv ads
y = number of radio ads
z = number of news ads


two formulas are indicated.


x + y + z = 88
2000x + 100y + 500z = 102400


they want as many tv ads as radio and news ads combined.
equation for that is x = y + z


since x = y + z, replace x with y + z in both equations to get;


y + z + y + z = 88
2000 * (y + z) + 100 * y + 500 * z = 102400


combine like terms and simplify to get:


2y + 2z = 88
2000y + 2000z + 100y + 500z = 102400


combine like terms again to get:


2y + 2z = 88
2100y + 2500z = 102400


multiply both sides of the first equation by 1050 and leave the second equation as is to get:


2100y + 2100z = 92400
2100y + 2500z = 102400


subtract the first equation from the second to get 400z = 10000


solve for z to get z = 25


since 2y + 2z = 88 and since y = 25, that equation becomes 2y + 50 = 88
subtract 50 from both sides to get 2y = 38
solve for z to get z = 19.


you have y = 19 and z = 25
since x = y + z, then x = 19 + 25 = 44


you have x = 44 and y = 19 and z = 25


go back to your original equations and replace x and y and z with their respective values to get:


x + y + z = 88 becomes 44 + 19  + 25 = 88 which becomes 88 = 88 which is true.
2000x + 100y + 500z = 102400 becomes 2000 * 44 + 100 * 19 + 500 * 25 = 102400 which becomes 102400 = 102400 which is true.


the values of x and y and z are confirmed to be good.


your solution is that 44 television ads and 19 radio ads and 25 news ads can be run each month.