Question

Statement: \[\dfrac{{2x}}{3} + 1 = \dfrac{{7x}}{{15}} + 3\] at \[x = 18\].
If this statement is true enter 1 if false enter 0. $ \\ $

Answer 1

Hint:
Linear equations are the equation whose variables have the highest exponential power as one, also known as the first-degree equation. A linear equation can have more than one variable.
In this question first, find the value of x by finding LCM of both sides of the equation and then RHS and LHS of the equation are cross multiplied after which value of x can be found, then compare the value of x obtained, and it is compared with the given value of the x.

Complete step by step solution:
\[\dfrac{{2x}}{3} + 1 = \dfrac{{7x}}{{15}} + 3\]
This can be written as
\[\dfrac{{2x}}{3} + \dfrac{1}{1} = \dfrac{{7x}}{{15}} + \dfrac{3}{1} - - (i)\]
We find the LCM for LHS and RHS of equation (i) separately
For LHS:
LCM \[(3,1) = 3\]
For RHS:
LCM \[(15,1) = 15\]
Hence we can write equation (i) as
\[
  \dfrac{{1 \times 2x + 3 \times 1}}{3} = \dfrac{{1 \times 7x + 15 \times 3}}{{15}} \\
  \Rightarrow \dfrac{{2x + 3}}{3} = \dfrac{{7x + 45}}{{15}} \\
 \]
Now cross multiply the obtained equation; hence we get
\[
  15\left( {2x + 3} \right) = 3\left( {7x + 45} \right) \\
  \Rightarrow 30x + 45 = 21x + 135 \\
 \]
Now solve for the value of x,
\[
  30x - 21x = 135 - 45 \\
 \Rightarrow 9x = 90 \\
 \Rightarrow x = \dfrac{{90}}{9} \\
 \Rightarrow x = 10 \\
 \]
Hence we get the value of \[x = 10\]

Now since the value of we obtained \[x = 10\] is not equal to \[x = 18\], hence we can say that
the given relation is false and we enter 0 as the input.
\[x = 10 \ne x = 18\]

Note: To find the value of a variable in an equation, always try to keep that variable in the Left-Hand Side of an equation. Cross multiplying can also be referred to as the method of removing the fractional terms in an equation by multiplying each term present in the denominator by each term present in the numerator.