Question

How do you solve \[k+3\dfrac{3}{4}=5\dfrac{2}{3}-1\dfrac{1}{3}?\]

Answer 1

Hint: We are asked to find the solution of \[k+3\dfrac{3}{4}=5\dfrac{2}{3}-1\dfrac{1}{3}.\] Firstly, we will learn what a linear equation in 1 variable term is. We use the hit and trial method to find the value of ‘k’. In this method, we put the value of k one by one by hitting arbitrary values and looking for the needed values. Another method is to apply algebra. We subtract the terms to get to our final term and get our required solution. We will also learn how to change the mix fraction into the proper fraction and then use it to simplify our given problem and solve it further.

Complete step-by-step solution:
In our given expression, we are asked to find the value of ‘k’ or we are asked how we will be able to solve this expression. We learn about the equation on one variable. One variable simply represents the equation that has one variable (say x, y or z) and another one constant. For example, x + 2 = 7 or 2(x + 2) = 7, etc.
Our equation is \[k+3\dfrac{3}{4}=5\dfrac{2}{3}-1\dfrac{1}{3}.\] Also it has just one variable k. We have to find the value of k which will satisfy our given equation. First, we will simplify our mixed fraction into a simple fraction and then we will solve by the method of hit and trial in which we will put a different value of ‘x’ and take which one fits the solution correctly. We have,
\[k+3\dfrac{3}{4}=5\dfrac{2}{3}-1\dfrac{1}{3}\]
Now,
\[3\dfrac{3}{4}=\dfrac{3\times 4+3}{4}\]
So, we get,
\[\Rightarrow 3\dfrac{3}{4}=\dfrac{12+3}{4}=\dfrac{15}{4}\]
Also,
\[5\dfrac{2}{3}=\dfrac{5\times 3+2}{3}\]
So, we get,
\[\Rightarrow 5\dfrac{2}{3}=\dfrac{17}{3}\]
And lastly, we have
\[1\dfrac{1}{3}=\dfrac{1\times 3+1}{3}\]
So, we get,
\[\Rightarrow 1\dfrac{1}{3}=\dfrac{4}{3}\]
Hence, our equation becomes
\[k+\dfrac{15}{4}=\dfrac{17}{3}-\dfrac{4}{3}\]
Let, k = 0, so, we get,
\[\Rightarrow 0+\dfrac{15}{4}=\dfrac{17}{3}-\dfrac{4}{3}\]
\[\Rightarrow \dfrac{15}{4}=\dfrac{17}{3}-\dfrac{4}{3}\]
On simplifying the right side, we get,
\[\Rightarrow \dfrac{15}{4}=\dfrac{13}{3}\]
which is not true.
So, k = 0 is not the solution.
Now, let k = 1 in \[k+\dfrac{15}{4}=\dfrac{17}{3}-\dfrac{4}{3}\] so we get,
\[\Rightarrow 1+\dfrac{15}{4}=\dfrac{17}{3}-\dfrac{4}{3}\]
On simplifying further, we get,
\[\Rightarrow \dfrac{19}{4}=\dfrac{13}{3}\]
which is incorrect.
Now, as we can see that this method is a bit hard and tricky, so we will use the method which involves algebraic operation and tools. With this tool, we will simplify and solve. So, as we have,
\[k+\dfrac{15}{4}=\dfrac{17}{3}-\dfrac{4}{3}\]
Now, firstly we will simplify the right side, so we get,
\[\Rightarrow k+\dfrac{15}{4}=\dfrac{17-4}{3}=\dfrac{13}{3}\]
So, we have,
\[\Rightarrow k+\dfrac{15}{4}=\dfrac{13}{3}\]
Now, we subtract \[\dfrac{15}{4}\] on both the sides so we get,
\[\Rightarrow k+\dfrac{15}{4}-\dfrac{15}{4}=\dfrac{13}{3}-\dfrac{15}{4}\]
In simplifying, we get,
\[\Rightarrow k=\dfrac{13\times 4-15\times 3}{12}\]
So, we get after solving,
\[\Rightarrow k=\dfrac{7}{12}\]
So, \[k=\dfrac{7}{12}\] is our required solution.

Note: Remember that we cannot add the variable to the constant. Usual mistakes like this where one adds the constants with the variables happen. For example, 3x + 6 = 9x, here one added 6 with 3 of x and made it 9x which is wrong. We cannot add constants and variables at once. Only the same variables are added to each other. We should be aware that when we change mix fraction into a simple fraction, we multiply the division with quotient and add the remainder and keep this in the numerator while the denominator stays the same. Also, when we usually open the bracket, we always multiply the term. With each term of the bracket that is a(b + c) = ab + ac.