AS - Maths (Simultaneous equation)
Sorry if i posted in the wrong section.
Need a step by step walk through for solving...
5x - 3y = 41
(7√2)x+(4√2)y=82
Need a step by step walk through for solving...
5x - 3y = 41
(7√2)x+(4√2)y=82
Why don't you make x the subject of the first equation and substitute it into the second?
2 ways, substitution or elimination. As you get older and wiser in maths, you'll use substitution more.
let x or y = whatever, so rearrange the top equation say x= (41-3y)/5, sub into second equation, solve for y, sub y back in to find x in either equation.
let x or y = whatever, so rearrange the top equation say x= (41-3y)/5, sub into second equation, solve for y, sub y back in to find x in either equation.
I tried that but im getting something wrong please point out what im doing wrong please. Answer should be y =5√2-7
5x= 41 + 3y
x = 41 + 3y /5
(7√2)(41+3y/5)+(4√2)y=82
(35√2)(41+3y)+(20√2)5y=410
1435√2+105y√2+100√2y=410
205y√2=410-1435√2
y=2-7√2
I know i've done something wrong and its probs cause im an idiot and struggling but a step by step solution would be very helpful.
5x= 41 + 3y
x = 41 + 3y /5
(7√2)(41+3y/5)+(4√2)y=82
(35√2)(41+3y)+(20√2)5y=410
1435√2+105y√2+100√2y=410
205y√2=410-1435√2
y=2-7√2
I know i've done something wrong and its probs cause im an idiot and struggling but a step by step solution would be very helpful.
Why did you multiply one side by 25 and the other by 5?
(edited 12 years ago)
Original post by Mr M
Why did you multiply one side by 25 and the other by 5?
I don't know im really struggling, i thought multiplying everything by 5 would get rid of the /5 cause i am meant to do it without a calculator.
I'm gonna be extra nice and run through the whole thing for you, bare with me. 
5x - 3y = 41
(7√2)x+(4√2)y=82
rearrange 5x-3y=41 into x=(41+3y)/5
sub x into 2nd equation:
(7√2)((41+3y)/5)+ (4√2)y=82
After some heavy algebra you should get to 287+61y=82
Then 61y=-205
y=-205/61
therefore 5x-3(-205/61)=41
5x= 41-(615/61)
x=((41-(615/61))/5)
5x - 3y = 41
(7√2)x+(4√2)y=82
rearrange 5x-3y=41 into x=(41+3y)/5
sub x into 2nd equation:
(7√2)((41+3y)/5)+ (4√2)y=82
After some heavy algebra you should get to 287+61y=82
Then 61y=-205
y=-205/61
therefore 5x-3(-205/61)=41
5x= 41-(615/61)
x=((41-(615/61))/5)
Original post by ashboot
I tried that but im getting something wrong please point out what im doing wrong please. Answer should be y =5√2-7
5x= 41 + 3y
x = 41 + 3y /5
(7√2)(41+3y/5)+(4√2)y=82
(35√2)(41+3y)+(20√2)5y=410
1435√2+105y√2+100√2y=410
205y√2=410-1435√2
y=2-7√2
I know i've done something wrong and its probs cause im an idiot and struggling but a step by step solution would be very helpful.
5x= 41 + 3y
x = 41 + 3y /5
(7√2)(41+3y/5)+(4√2)y=82
(35√2)(41+3y)+(20√2)5y=410
1435√2+105y√2+100√2y=410
205y√2=410-1435√2
y=2-7√2
I know i've done something wrong and its probs cause im an idiot and struggling but a step by step solution would be very helpful.
Your x is incorrect, you need to divide everything on the RHS by 5.
I'm sorry for the heavy fractions I've left, I would simplify more if I had a calculator.
Original post by ashboot
I don't know im really struggling, i thought multiplying everything by 5 would get rid of the /5 cause i am meant to do it without a calculator.
Multiplying by 5 does get rid of the /5 but you multiplied by 25 because you changed 7 to 35 as well. (You multiplied by 5 twice).
Original post by ashboot
I think im just completely defeated by this question tbh.
As an example of where you went wrong:
Quick Reply
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