How do you solve \[5x + y = 1\] and \[3x + 2y = 2\] using substitution?

Hint: To solve the given simultaneous equation, substitute the given equation for \[x\] of equation 2 in equation 1 and hence simplify the terms by substituting the equation to get the value of \[y\] . Substitute the value obtained of \[y\]in equation 2, hence the value of \[x\] is obtained.

Complete step-by-step answer:
Let us write the given equation
\[5x + y = 1\] …………………………. 1
\[3x + 2y = 2\] ..………………………… 2
The standard form of simultaneous equation is
\[Ax + By = C\]
Equation 1 can be written in terms of \[y\] as
\[\Rightarrow y = 1 - 5x\]
Hence, substitute the value of \[y\] in equation 2 in terms of \[x\] as
\[\Rightarrow 3x + 2y = 2\]
\[\Rightarrow 3x + 2\left( {1 - 5x} \right) = 2\]
After substituting the y term, simplify the obtained equation
\[\Rightarrow 3x + 2 - 10x = 2\]
\[\Rightarrow - 7x = 2 - 2\]
\[ \Rightarrow - 7x = 0\]
Which implies that
\[\Rightarrow x = 0\]
As we got the value of \[x\], substitute the value of \[x\] as 0 in equation 1 we get,
\[\Rightarrow 5x + y = 1\]
\[\Rightarrow 5\left( 0 \right) + y = 1\]
\[y = 1\]
Therefore, the value of \[y\] is 1.
Hence the values of \[x\] and \[y\] are
\[x = 0\] and \[y = 1\]

Additional Information:
Simultaneous equations are a set of two or more equations, each containing two or more variables whose values can simultaneously satisfy both or all the equations in the set, the number of variables being equal to or less than the number of equations in the set.
The method of solving "by substitution" works by solving one of the equations for one of the variables, and then plugging this back into the other equation, "substituting" for the chosen variable and solving for the other.


Note: We know that Simultaneous equations are two equations, each with the same two unknowns and are "simultaneous" because they are solved together, hence the key point to solve this kind of equations is we need to combine all the terms and then simplify the terms to get the value of \[x\] also the value of \[y\].