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Here we have been given with two equations and we will solve them by substitution by elimination method.
Transposing y in 1st equation from L.H.S. to R.H.S. it would be negative,
[tex] \implies \: \displaystyle\sf{x \: = \: 6 - y}[/tex]
Now, we would substitute this value of x in Equation No 1.
[tex] \implies \: \displaystyle\sf{2 \: (6 - y) \: = \:15 }[/tex]
[tex] \implies \: \displaystyle\sf{2 \: \times (6 - y) \: = \:15 }[/tex]
[tex]\implies \: \displaystyle\sf{12 - 2y \: = \:15 }[/tex]
Transposing 12 to R.H.S.,
[tex] \implies \: \displaystyle\sf{ - 2y \: = \:15 - 12 }[/tex]
[tex]\implies \: \displaystyle\sf{ - 2y \: = \:3}[/tex]
[tex]\implies \: \displaystyle\bf{ y \: = \: - \dfrac{3}{2} }[/tex]
Therefore, value of y is - 3/2.
★ Finding out value of x:-
[tex] \implies \: \displaystyle\sf{x \: = \: 6 - y }[/tex]
[tex]\implies \: \displaystyle\sf{x \: = \: 6 - ( \dfrac{ - 3}{ 2} )}[/tex]
[tex]\implies \: \displaystyle\sf{x \: = \: \frac{12 + 3}{2} }[/tex]
[tex]\implies \: \displaystyle\bf{x \: = \: \frac{15}{2} }[/tex]
Henceforth,