Solve the following equation for $x$, and express its value in terms of $c$.
$3(c x-7)=9$
The value of $x$ in the equation is $\square$.
Answer (1)
Distribute the constant: 3 ( c x − 7 ) = 3 c x − 21 .
Add 21 to both sides: 3 c x = 30 .
Divide by 3 c : x = 3 c 30 .
Simplify the fraction: x = c 10 .
c 10
Explanation
Understanding the Problem We are given the equation 3 ( c x − 7 ) = 9 and asked to solve for x in terms of c . This means we want to isolate x on one side of the equation and have an expression involving c on the other side.
Distributing the Constant First, distribute the 3 on the left side of the equation: 3 ( c x − 7 ) = 3 c x − 21
So the equation becomes: 3 c x − 21 = 9
Isolating the x Term Next, add 21 to both sides of the equation to isolate the term with x :
3 c x − 21 + 21 = 9 + 21
3 c x = 30
Solving for x Now, divide both sides of the equation by 3 c to solve for x :
3 c 3 c x = 3 c 30
x = 3 c 30
Simplifying the Fraction Finally, simplify the fraction by dividing both the numerator and the denominator by 3 :
x = 3 c ÷ 3 30 ÷ 3
x = c 10
Final Answer Therefore, the value of x in terms of c is c 10 .
Examples
In physics, if you have a formula relating distance, speed, and time, like d = c \t , where d is the distance, c is the speed (which is constant), and t is the time, and you know the distance and the product of speed and another variable x (where c x = k ), you can use this problem's approach to find how the variable x relates to the time t . For instance, if 3 ( c x − 7 ) = 9 represents a specific condition in the problem, solving for x in terms of c helps you understand the relationship between these variables and how they affect the time or distance in the physical scenario. This kind of algebraic manipulation is crucial for solving many physics problems.
