Question

How do you solve 35 = – 5 + 2x – 7x?

Answer 1

Hint: We are asked to find the solution of 35 = – 5 + 2x – 7x. Firstly, we learn what a linear equation is in 1 variable term. We will use a hit and trial method to find the value of ‘x’. In this method, we put the value of ‘x’ one by one by hitting arbitrary values and looking for the needed values. Once we work with the hit and trial method we will try another method where we apply algebra. We subtract, add, or multiply the terms to get to our final term and get our required solution.

Complete step by step answer:
We are given that we have 35 = – 5 + 2x – 7x. We are asked to find the value of ‘x’, or we are asked how we will be able to solve this expression. We will 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 = 4, 2 – x = 2, 2x, 2y, etc. Our equation 35 = – 5 + 2x – 7x also has just one variable ‘x’. We have to find the value of ‘x’ which will satisfy our given equation. Firstly, we will try the method of hit and trial. In which we will put a different value of ‘x’ and take which one fits the solution correctly.
Let x = 0, so putting x = 0 in 35 = – 5 + 2x – 7x, we get,
\[\Rightarrow 35=-5+2\times 0-7\times 0\]
\[\Rightarrow 35=-5\]
which is not true. So, x = 0 is not the solution to our problem.
Let x = 1, so putting x = 1 in 35 = – 5 + 2x – 7x, we get,
\[\Rightarrow 35=-5+2\times 1-7\times 1\]
\[\Rightarrow 35=-5+2-7\]
On simplifying, we get,
\[\Rightarrow 35=-10\]
which is not true. So, x = 1 is not the solution to our problem.
Let x = – 5, so putting x = – 5 in 35 = – 5 + 2x – 7x, we get,
\[\Rightarrow 35=-5+2\times \left( -5 \right)-7\times \left( -5 \right)\]
We get,
\[\Rightarrow 35=20\]
which is not true. So, x = – 5 is not the solution to our problem.
Let x = – 8, so putting x = – 8 in 35 = – 5 + 2x – 7x, we get,
\[\Rightarrow 35=-5+2\times \left( -8 \right)-7\times \left( -8 \right)\]
We get,
\[\Rightarrow 35=-5+40\]
\[\Rightarrow 35=35\]
which is true. So, x = – 8 is the solution to our problem.
Now, we will find the solution by another method in which we will use algebraic operation. Now we have,
\[35=-5+2x-7x\]
Add 5 on both the sides, we get,
\[\Rightarrow 35+5=-5+5+2x-7x\]
On simplifying, we get,
\[\Rightarrow 40=-5x\]
Dividing both the sides by – 5, we get,
\[\Rightarrow \dfrac{40}{-5}=\dfrac{-5x}{-5}\]
So, we get,
\[\Rightarrow x=-8\]
So, x = – 8 is our required solution.

Note:
 Remember that, we cannot add the variable to the constant. Usual mistakes like this where one adds constants with variables usually happen. For example, 3x + 6 = 9x, here one added ‘6’ with 3 of x made it 9x, this is wrong, we cannot add constant and variable at once. Only the same variables are added to each other. Remember when we divide a positive term by a negative value the solution we get is a negative term this may happen that we skip – sign.