
How do you solve $c+(4-3c)-2=0$?
Answer
552.6k+ views
Hint: The equation given in the above question is a linear equation in one variable, that is c. The question says that we have to solve the given equation in c. In other words, we have to find the values of c which will satisfy the given equation. Try to find the value of c by performing some mathematical operations.
Complete step by step solution:
The given equation says that $c+(4-3c)-2=0$ …. (i)
To find the value of c that satisfies the above given equation we shall analyse the given equation and then figure out to find the required value of c. In equation (i), we can use a bracket with some inside it, on the left hand side of the equation. Let the first open up that bracket. When we open up that bracket the equation will change to $c+4-3c-2=0$.
The thing that we can next do is that we can club the terms that are multiples of c on one side of the equation (let us say left hand side) and we can club the constant terms on the other side of the equation (i.e. right hand side).When we do this the equation changes to $c-3c=2-4$Then, the further calculation gives us that $-2c=-2$. Then this means that $c=1$.
Therefore, $c=1$ is the value of c that satisfies the given equation.
Note: There are many ways in which we can solve this question. We can also directly add and subtract terms such that the left hand side is left with only terms that are multiples of c and the right hand side is left with only constant terms.However, students must note that the number of solutions to a given equation in one variable is always less than or equal to the degree of the polynomial in the given equation.
Complete step by step solution:
The given equation says that $c+(4-3c)-2=0$ …. (i)
To find the value of c that satisfies the above given equation we shall analyse the given equation and then figure out to find the required value of c. In equation (i), we can use a bracket with some inside it, on the left hand side of the equation. Let the first open up that bracket. When we open up that bracket the equation will change to $c+4-3c-2=0$.
The thing that we can next do is that we can club the terms that are multiples of c on one side of the equation (let us say left hand side) and we can club the constant terms on the other side of the equation (i.e. right hand side).When we do this the equation changes to $c-3c=2-4$Then, the further calculation gives us that $-2c=-2$. Then this means that $c=1$.
Therefore, $c=1$ is the value of c that satisfies the given equation.
Note: There are many ways in which we can solve this question. We can also directly add and subtract terms such that the left hand side is left with only terms that are multiples of c and the right hand side is left with only constant terms.However, students must note that the number of solutions to a given equation in one variable is always less than or equal to the degree of the polynomial in the given equation.
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