SOLUTION: i have to 2 equations x+y=20 and 25x+15y=430. the end goal is to find how many of x and y there is. i think i need to cancel out the 15 but i dont know where to start.
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Question 1055734: i have to 2 equations x+y=20 and 25x+15y=430. the end goal is to find how many of x and y there is. i think i need to cancel out the 15 but i dont know where to start. Found 2 solutions by Boreal, MathTherapy:Answer by Boreal(15235) (Show Source):
You can put this solution on YOUR website! x+y=20
25x+15y=430.
Multiply the top equation by -15
-15x-15y=-300
25x+15y=430
add them and the y disappear
10x=130
divide by 10
x=13
Put into first equation and 13+y=20; y=7
Check with the second equation
25(13)+15(7)=325+105=430
(13,7)
Note, you could multiply the first equation by -25, and then when you added, you would eliminate the x s and get the y. You eliminate whatever appears to be easier.
You can put this solution on YOUR website! i have to 2 equations x+y=20 and 25x+15y=430. the end goal is to find how many of x and y there is. i think i need to cancel out the 15 but i dont know where to start.
x + y = 20 ------- eq (i)
25x + 15y = 430 ------- eq (ii)
You can multiplying eq (i) by - 15 to ELIMINATE y and find the value of x
Then substitute value of x into eq (i) (easier of the 2) to get value of y
OR
Divide eq (ii) by 5 to get: 5x + 3y = 86 ------ eq (ii).
Then multiply eq (i) by - 3 to get eq (iii), which you then add to eq (ii) to ELIMINATE y and find the value of x.
You'd have 1 more equation to change but after all, you'd be dealing with smaller numbers. It's your choice though.
