# simultaneous equationsWatch

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#1
Solve the simultaneous equations 7x - (3 √50 ) y = 9√5 and (2√5)x + y =34.
0
13 years ago
#2
(Original post by e-n-i-g-m-a)
Solve the simultaneous equations 7x - (3 √50 ) y = 9√5 and (2√5)x + y =34.
Where are you getting stuck? Can you find an expression for y in terms of x using one of the equations and from there substitute this into the other (so obtaining a single equation in x that you can solve) ?
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#3
(Original post by Gaz031)
Where are you getting stuck? Can you find an expression for y in terms of x using one of the equations and from there substitute this into the other (so obtaining a single equation in x that you can solve) ?

well ya i can figure out that y = 34 - (2√5)x but i dont know wat to do after i substitute this equation in the other one .. i mean how do i multilply this value of y with (3√5)y in the first equation!!!
0
13 years ago
#4
(Original post by e-n-i-g-m-a)
Solve the simultaneous equations 7x - (3 √50 ) y = 9√5 and (2√5)x + y =34.
7x - 3y√50 = 9√5
2x√5 + y = 34

y = 34 - 2x√5

7x - (34 - 2x√5)(3√50) = 9√5
7x - (102√50 - 6x√250) = 9√5
7x - (510√2 - 30x√10) = 9√5
7x + 30x√10 = 9√5 + 510√2
x = (9√5 + 510√2)/(7 + 30√10)

y = 34 - 2x√5
y = 34 - 2(9√5 + 510√2)√5/(7 + 30√10)

wow messy... havent done maths for 3 months or so..
0
13 years ago
#5
(Original post by e-n-i-g-m-a)
Solve the simultaneous equations 7x - (3 √50 ) y = 9√5 and (2√5)x + y =34.
7x = 9rt5 + 3rt50y
-> x = (9rt5 + 3rt50y)/7

2rt5x = 34 - y
-> x = (34 - y)/(2rt5)

-> (34 - y)/(2rt5) = (9rt5 + 3rt50y)/7
-> 238 - 7y = 90 + 30rt10y
-> 148 = 30rt10y + 7y
-> y(30rt10 + 7) = 148
-> y = 1.45 (3.S.F)

-> x = (9rt5 + 3rt50*1.453)/7
-> x = 7.28 (3.S.F)
0
13 years ago
#6
yea the answers are x= 7.278(3dp) or x= ((9rt5)+(510rt2))/(7+(30rt10)) in surd form and y= 1.453 or y = 148/(7+(30rt10)) in surd form. The question requires manipulation of surds. As long as you can play around with big numbers then you should be able to solve it. You just make the second equation which has a lone y value equal to the rest of the equation and substitute that into the 1st equation. You need to simplify the surds and rationalise the denominators in some cases just to make it a bit easier. Perhaps you should start by simplifying the 3rt50 into 15rt2 and then work from there. I hope I've explained it properly.

Angu
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