solve the system: x - 3y = 5 2x + y = 4
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x - 3y = 5 ........(1)
2x + y = 4 ........(2)
using the eliminattion methos, multiply (2)...
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We'll try to solve the system using another method: substitution method.
x - 3y = 5
x = 5 + 3y (1)
We'll substitute (1) in the second equation of the system:
2x + y = 4
2(5 + 3y) + y = 4
We'll remove the brackets:
10 + 6y + y = 4
We'll combine like terms:
10 + 7y = 4
We'll subtract 10:
7y = -6
We'll divide by 7:
y = -6/7
We'll substitute y in (1):
x = 5 - 18/7
x = (35-18)/7
x = 17/7
The solution of the system is {(17/7;-6/7)}
x-3y =5 ..(1) and 2x+y =4....(2)
We eliminate x by (2)-2(1):
y- -6y = 4-2*5 = =-6
7y = -6
y = -6/7 .
Substituting y = -6/7 in (1) : x - 3y = 5.
x= 5+3y = 5+3(-6/7) = 17/7
x= 17/7 and y =-6/7.
(2x+y)-2(x-3y) = 4-2*5
Given the two equations:
x - 3y = 5 ...(1)
2x + y = 4 ...(2)
(2) - 2* (1)
=> 2x + y - 2* (x - 3y) = 4 - 10
=> 2x + y - 2x + 6y = -6
=> 7y = -6
=> y = -6/7
susbstituting y = -6/7 in x - 3y = 5
=> x = 5 + 3*(-6 /7)
=> x = 5 - 18 /7
=> x = 17 / 7
Therefore x is 17 / 7 and y is -6/7
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