Question

solve the equation by substitution method

x + y = 2m
mx - ny = m² + n²

Answer 1

From 1st eqn
X= 2m-y
mx-ny=m^2 +n^2
m(2m-y) -ny= m^2+n^2
2m^2 - ym - ny = m^2 + n^2
-(m+n) y= m^2+n^2-2m^2
= n^2- m^2
= -(m^2- n^2)
( m+n) y = (m+n)(m-n)
Y= m-n.

X= 2m-y
= 2 m-(m-n)
= 2m-m+n
=m+n.

Answer 2

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given equations =>

$x + y = 2m \: \: \: ......eq _{1}\\ \\ mx - ny = {m}^{2} + {n}^{2} \: \: \: ......eq _{2} \\ \\ from \: eq _{1} \\ \\ x = 2m - y \: \: \: .....eq _{3} \\ \\ plug \: the \: value \: of \: x \: in \: eq _{2} \\ \\ then \: we \: get \\ \\ m(2m - y) - ny = {m}^{2} + {n}^{2} \\ \\ 2 {m}^{2} - my - ny = {m}^{2} + {n}^{2} \\ \\ {m}^{2} - {n}^{2} - y(m + n) = 0 \\ \\ ( {m}^{2} - {n}^{2} ) - y(m + n) = 0 \\ \\ - y(m + n) = - ( {m}^{2} - {n}^{2} ) \\ \\ y(m + n) = (m + n)(m - n) \\ \\ we \: know \: = > ( { \alpha }^{2} - { \beta }^{2} ) = ( \alpha + \beta )( \alpha - \beta ) \\ \\ y \: = m - n \: \: \: \: ......answer \\ \\ now \\ \\ plug \: the \: value \: of \: y \: in \: eq _{1} \: we \: get \: \\ \\ x + y \: = \: 2m \\ \\ x + (m - n) = 2m \\ \\ x \: \: = \: 2m - m + n \\ \\ x \: = \: m + n \: \: \: \: ......answer$

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