- Formula Clearance
- 5 formula clearance exercises
- First exercise
- Solution
- Second exercise
- Solution
- Third exercise
- Solution
- Fourth exercise
- Solution
- Fifth exercise
- Solution
- References
The solved exercises clearance formulas allow us to understand better this operation. Formula clearing is a widely used tool in mathematics.
Solving for a variable means that the variable must be left on one side of equality, and everything else must be on the other side of equality.
When you want to clear a variable, the first thing to do is take everything that is not said variable to the other side of equality.
There are algebraic rules that must be learned in order to isolate a variable from an equation.
Not all formulas can solve for a variable, but this article will present exercises where it is always possible to solve for the desired variable.
Formula Clearance
When you have a formula, you first identify the variable. Then all the addends (terms that are added or subtracted) are passed to the other side of the equality by changing the sign of each addend.
After passing all the addends to the opposite side of the equality, it is observed if there is any factor multiplying the variable.
If yes, this factor must be passed to the other side of equality by dividing the entire expression on the right and keeping the sign.
If the factor is dividing the variable, then this must be passed by multiplying the entire expression on the right, keeping the sign.
When the variable is raised to some power, for example "k", a root with index "1 / k" is applied to both sides of the equality.
5 formula clearance exercises
First exercise
Let C be a circle such that its area is equal to 25π. Calculate the radius of the circumference.
Solution
The formula for the area of a circle is A = π * r². Since we want to know the radius, then we proceed to clear «r» from the previous formula.
As there are no adding terms, we proceed to divide the factor «π» that is multiplying «r²».
We then obtain r² = A / π. Finally, we proceed to apply a root with index 1/2 to both sides and we will obtain r = √ (A / π).
Substituting A = 25, we get that r = √ (25 / π) = 5 / √π = 5√π / π ≈ 2.82.
Second exercise
The area of a triangle is equal to 14 and its base is equal to 2. Calculate its height.
Solution
The formula for the area of a triangle is equal to A = b * h / 2, where "b" is the base and "h" is the height.
As there are no terms adding to the variable, we proceed to divide the factor «b» that is multiplying «h», from which it follows that A / b = h / 2.
Now the 2 that is dividing the variable is passed to the other side by multiplying, so that it turns out that h = 2 * A / h.
Substituting A = 14 and b = 2 we get that the height is h = 2 * 14/2 = 14.
Third exercise
Consider the equation 3x-48y + 7 = 28. Solve for the variable «x».
Solution
When observing the equation, two addends can be seen next to the variable. These two terms must be passed to the right side and their sign changed. So you get
3x = + 48y-7 + 28 ↔ 3x = 48y +21.
Now we proceed to divide the 3 that is multiplying the «x». Therefore, it follows that x = (48y + 21) / 3 = 48y / 3 + 27/3 = 16y + 9.
Fourth exercise
Solve for the variable «y» from the same equation from the previous exercise.
Solution
In this case the addends are 3x and 7. Therefore, when passing them to the other side of the equality we have that -48y = 28 - 3x - 7 = 21 - 3x.
The '48 is multiplying the variable. This is passed to the other side of equality by dividing and preserves the sign. Therefore, we obtain:
y = (21-3x) / (- 48) = -21/48 + 3x / 48 = -7/16 + x / 16 = (-7 + x) / 16.
Fifth exercise
It is known that the hypotenuse of a right triangle is equal to 3 and one of its legs is equal to √5. Calculate the value of the other leg of the triangle.
Solution
The Pythagorean theorem says that c² = a² + b², where "c" is the hypotenuse, "a" and "b" are the legs.
Let "b" be the leg that is not known. Then you start by passing «a²» to the opposite side of the equality with the opposite sign. In other words, we obtain b² = c² - a².
Now the root «1/2» is applied to both sides and we obtain that b = √ (c² - a²). By substituting the values of c = 3 and a = √5 we obtain that:
b = √ (3²- (√5) ²) = √ (9-5) = √4 = 2.
References
- Fuentes, A. (2016). BASIC MATH. An Introduction to Calculus. Lulu.com.
- Garo, M. (2014). Mathematics: quadratic equations: How solve a quadratic equation. Marilù Garo.
- Haeussler, EF, & Paul, RS (2003). Mathematics for management and economics. Pearson Education.
- Jiménez, J., Rofríguez, M., & Estrada, R. (2005). Math 1 SEP. Threshold.
- Preciado, CT (2005). Mathematics Course 3rd. Editorial Progreso.
- Rock, NM (2006). Algebra I Is Easy! So Easy. Team Rock Press.
- Sullivan, J. (2006). Algebra and Trigonometry. Pearson Education.