How does one solve by substitution $y = 3x + 5$ and $4x - y = 5$ ?

Hint: We can solve this particular question by substitution as we can substitute the expression for $y$ that is $y = 3x + 5$ given to us by the primary equation into the second equation that is $4x - y = 5$ . Once you find $x$you can easily get $y$ .

Complete solution step by step:
Let us consider ,
$y = 3x + 5..................(1)$
And
$4x - y = 5...................(2)$
We can substitute the expression for $y$ given to us by the primary equation into the second equation:
Take equation $(2)$ that is ,
$4x - y = 5$
Now substitute $y = 3x + 5..................(1)$ in $(2)$ as follow ,
$
\Rightarrow 4x - y = 5 \\
\Rightarrow 4x - (3x + 5) = 5 \\
\Rightarrow 4x - 3x - 5 - 5 = 0 \\
\Rightarrow x - 10 = 0 \\
\Rightarrow x = 10 \\
$
Then substitute this value of $x$ for $x$ within the first equation:
$
\Rightarrow y = 3x + 5 \\
\Rightarrow y = 3(10) + 5 \\
\Rightarrow y = 35 \\
$
Additional Information: Let us have three equations and three variables $x,y,z$ . Now develop an equation with $x$and segregate it say $x$ in terms of $y,z$ . once we put this value of $x$ in two other equations we get two equations in $y,z$ . We can now find $y$ in terms of $z$ say using the second equation and after we put in the third equation we get the value of $z$ . Once $z$ is found , it's easy to search out $y$ so $x$ .

Note: Well, I might say that it is easier after you have few equations and variables. If you have got two equations and a pair of variables it is ok; once you get to three equations and three variables it becomes more complicated, it's still possible, but you#39;ve got more work to try and do. the quantity of substitutions increases along with the chance to create mistakes. More than three equations and three variables and it gets almost impossible and other methods would be better.