Form the pair of linear equations for the following problems and find their solution by substitution method.(i) The difference between two numbers is 26 and one number is three times the other. Find them.(ii) The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.
Form the pair of linear equations for the following problems and find their solution by substitution method.(i) The difference between two numbers is 26 and one number is three times the other. Find them.(ii) The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.

Arrangement (i):

Leave the two numbers alone x and y individually, to such an extent that y > x.

As indicated by the inquiry,

\[y\text{ }=\text{ }3x\text{ }\ldots \text{ }\ldots \text{ }\ldots \text{ }\left( 1 \right)\]

\[y\text{ }-x\text{ }=\text{ }26\text{ }\ldots \text{ }\ldots \text{ }..\left( 2 \right)\]

Subbing the worth of (1) into (2), we get

\[3x\text{ }-x\text{ }=\text{ }26\]

\[x\text{ }=\text{ }13\text{ }\ldots \text{ }\ldots \text{ }\ldots \text{ }.\text{ }\left( 3 \right)\]

Subbing (3) in (1), we get \[y\text{ }=\text{ }39\]

Consequently, the numbers are 13 and 39.

(ii) The bigger of two valuable points surpasses the more modest by 18 degrees. Discover them.

Arrangement (ii) :

Let the bigger point by \[{{x}^{o}}\]  and more modest point be \[{{y}^{o}}\] .

We realize that the amount of two valuable pair of points is consistently \[{{180}^{o}}\] .

As per the inquiry,

\[x\text{ }+\text{ }y\text{ }=\text{ }{{180}^{o}}\ldots \ldots \ldots \ldots \ldots .\text{ }\left( 1 \right)\]

\[x\text{ }-y\text{ }=\text{ }{{18}^{o~}}\ldots \ldots \ldots \ldots \ldots ..\left( 2 \right)\]

From (1), we get \[x\text{ }=\text{ }{{180}^{o}}~-y\text{ }\ldots \ldots \ldots \ldots .\text{ }\left( 3 \right)\]

Subbing (3) in (2), we get

\[{{180}^{o~}}-y\text{ }-y\text{ }={{18}^{o}}\]

\[{{162}^{o}}~=\text{ }2y\]

\[y\text{ }=\text{ }{{81}^{o}}~\ldots \ldots \ldots \ldots ..\text{ }\left( 4 \right)\]

Utilizing the worth of y in (3), we get

\[x\text{ }=\text{ }{{180}^{o}}~-{{81}^{o}}\]

\[=\text{ }{{99}^{o}}\]

Thus, the points are \[\mathbf{9}{{\mathbf{9}}^{\mathbf{o}}}~\mathbf{and}\text{ }\mathbf{8}{{\mathbf{1}}^{\mathbf{o}}}\] .