How do you solve \[2x - 4 = 5x - 19\] ?
Answer
Verified
443.7k+ views
Hint:We are given the equation with one variable. We are about to solve the equation that is to find the value of that variable in the equation. For that first we will try to separate the variable terms and constant terms. Then we will perform the necessary mathematical operations to get the value of the variable.
Complete step by step answer:
Given that,
\[2x - 4 = 5x - 19\]
Now \[x\] is the variable. So let’s first separate the constants and variables. Taking \[2x\] on RHS and 19 on LHS of the equation we get,
\[19 - 4 = 5x - 2x\]
Now perform the mathematical operations,
\[15 = 3x\]
Now divide 15 by 3 to get the value of variable,
\[\dfrac{{15}}{3} = x\]
Thus
\[5 = x\]
This is our final answer.
Additional information:
Variables are having different values. Variables are the English alphabets that are assigned to different values. Constants have fixed value. Constants are the numbers.
Note:
Given \[2x - 4 = 5x - 19\]
Now we will move variables on LHS and constants on RHS.
\[2x - 5x = - 19 + 4\]
\[ - 3x = - 15\]
Now cancelling the minus sign from both sides,
\[x = \dfrac{{15}}{3} = 5\]
Answer remains the same.
Here note that when terms shift from one side of the equation to the other side they change their sign. That is a positive term becomes negative and negative becomes positive. This is the most important step to remember because if we don’t change the sign the answer differs and rather it is totally wrong. Operations on variables are the same as constants only the variable is attached to the terms.
Complete step by step answer:
Given that,
\[2x - 4 = 5x - 19\]
Now \[x\] is the variable. So let’s first separate the constants and variables. Taking \[2x\] on RHS and 19 on LHS of the equation we get,
\[19 - 4 = 5x - 2x\]
Now perform the mathematical operations,
\[15 = 3x\]
Now divide 15 by 3 to get the value of variable,
\[\dfrac{{15}}{3} = x\]
Thus
\[5 = x\]
This is our final answer.
Additional information:
Variables are having different values. Variables are the English alphabets that are assigned to different values. Constants have fixed value. Constants are the numbers.
Note:
Given \[2x - 4 = 5x - 19\]
Now we will move variables on LHS and constants on RHS.
\[2x - 5x = - 19 + 4\]
\[ - 3x = - 15\]
Now cancelling the minus sign from both sides,
\[x = \dfrac{{15}}{3} = 5\]
Answer remains the same.
Here note that when terms shift from one side of the equation to the other side they change their sign. That is a positive term becomes negative and negative becomes positive. This is the most important step to remember because if we don’t change the sign the answer differs and rather it is totally wrong. Operations on variables are the same as constants only the variable is attached to the terms.
Recently Updated Pages
Glucose when reduced with HI and red Phosphorus gives class 11 chemistry CBSE
The highest possible oxidation states of Uranium and class 11 chemistry CBSE
Find the value of x if the mode of the following data class 11 maths CBSE
Which of the following can be used in the Friedel Crafts class 11 chemistry CBSE
A sphere of mass 40 kg is attracted by a second sphere class 11 physics CBSE
Statement I Reactivity of aluminium decreases when class 11 chemistry CBSE
Trending doubts
10 examples of friction in our daily life
The correct order of melting point of 14th group elements class 11 chemistry CBSE
Difference Between Prokaryotic Cells and Eukaryotic Cells
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE
State and prove Bernoullis theorem class 11 physics CBSE
What organs are located on the left side of your body class 11 biology CBSE