Question

How do you solve \[y = 4x + 45\] and \[x = 4y\]?

Answer 1

Hint: Here we have a system of two linear equations with two variables. We need to find the value of ‘x’ and ‘y’. We solve this using a substitution method. First we need to solve one equation for one of the variables and then we need to substitute this expression into another equation and we solve it. Using this we will have one variable value and to find the other we substitute the obtained variable value in any one of the given equations.

Complete step by step answer:
Given,
\[y = 4x + 45{\text{ }} - - - (1)\]
\[x = 4y{\text{ }} - - - (2)\]
From equation (2) we have,
\[x = 4y\]
Now we substitute this ‘x’ value in equation (1) we have,
\[y = 4(4y) + 45\]
\[y = 16y + 45\]
Thus we have a linear equation with one variable and we can simplify for ‘y’,
\[y - 16y = 45\]
\[ - 15y = 45\]
Divide the whole equation by -15 we have,
\[y = - \dfrac{{45}}{{15}}\]
\[ \Rightarrow y = - 3\].
To find the value of ‘x’ we need to substitute the obtained ‘y’ value in any one of the equations. Let’s substitute the value of ‘y’ in equation (2) we have,
\[x = 4y\]
\[x = 4( - 3)\]
\[ \Rightarrow x = - 12\].

Thus we have the solution is \[ \Rightarrow x = - 12\] and \[ \Rightarrow y = - 3\].

Note: To check whether the obtained answer is correct or not, we substitute the obtained value in the given problem,
\[x = 4y\]
\[ - 12 = 4( - 3)\]
\[ \Rightarrow - 12 = - 12\]
That is equation 2 satisfies, similarly it will also satisfy equation (1). Hence the obtained answer is correct. We can also solve the given problem by using elimination method or by cross product method.