MHB Substitution Method to solve linear simultaneous equation

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The discussion focuses on solving a system of linear equations using the substitution method. The user transformed the equations into a common denominator and derived two equations: 5y - 5x = 1 and 5y + 2x = 5. After substituting, they found x = 4/7, but expected the solution to be x = 0. A suggestion was made to multiply the first equation by 6 and simplify to potentially correct the solution. The conversation highlights the importance of careful manipulation of equations in the substitution method.
Yazan975
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What I have done:

I changed all fractions to common denom and that gave me

5y-5x=1 (1) *I numbered the fractions
5y+2x=5 (2)

Then: 5y=5-2x

Substitute into equation 1
(5-2x)-5x=1
5-7x=1
x=4/7

Thing is my answer says I should be getting x=0

Any hints?
 

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When you multiply the first equation by $6$ and simplify, you should get $y-5x=1$.
 
Thread 'Erroneously finding discrepancy in transpose rule'

Thread 'Erroneously finding discrepancy in transpose rule'

Obviously, there is something elementary I am missing here. To form the transpose of a matrix, one exchanges rows and columns, so the transpose of a scalar, considered as (or isomorphic to) a one-entry matrix, should stay the same, including if the scalar is a complex number. On the other hand, in the isomorphism between the complex plane and the real plane, a complex number a+bi corresponds to a matrix in the real plane; taking the transpose we get which then corresponds to a-bi...
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