A kosarad üres
Solving equations with elimination,
with the “balance method”
with the “balance method”
While solving the equation it is usually necessary to transform it too. This process is called rearrangement of the equation. The transformations, in which we do not lose any of the roots of the equation and we do not get such solutions which are not roots of the original equation (these are called false roots), are called equivalent transformations.
During the rearrangement we act according to the rules resulting from the balance method. It means that during equivalent transformations we do the same operations on both sides of the equation taking the following conditions into consideration
- We can add or subtract the same number to or from both sides of the equation:
\latex{ 6x - 3 = 9 \;\;/+3},
\latex{ 6x = 12. }
\latex{ 6x = 12. }
- We can multiply or divide both sides of the equation by the same number different from \latex{0}:
\latex{ 6x = 12\;\; / \div6, }
\latex{ x = 2. }
\latex{ x = 2. }
- We can add or subtract a term containing an unknown if it does not change the fundamental set of the equation:
\latex{ 8x = 6x - 2\;\; /-6x, }
\latex{ 2x = -2, }
\latex{ x = -1. }
\latex{ 2x = -2, }
\latex{ x = -1. }
- If we multiply by an expression containing an unknown, then we might get false roots, and if we divide by it, then it might lead to loss of root. Therefore we preferably should not apply this operation. If it still happens, then we should pay extra attention to the checking of solutions of the equation!
Example 1
I thought of a number, I subtracted six from its double, I halved the result, and thus finally I got three. Which number did I think of?
Solution I
We can get the solution of the exercise, if by elimination we trace back the resulting numbers in each step.
In the last step when dividing by \latex{ 2 } we got \latex{ 3 } as a result, so the dividend was \latex{ 6 }. \latex{ 6 } resulted so that we subtracted \latex{ 6 } from a number, so the minuend was \latex{ 12 }. \latex{ 12 } is the double of the number thought of, thus the number though of was \latex{ 6 }.
Indeed: \latex{\frac{2\times 6-6}{2} =\frac{12-6}{2} =\frac{6}{2} =3}
Indeed: \latex{\frac{2\times 6-6}{2} =\frac{12-6}{2} =\frac{6}{2} =3}
Solution II
Let us solve the exercise with an equation. Let the number thought of be x. According to the text of the exercise we can set up the following equation:
\latex{\frac{2x-6}{2} =3}
\latex{\frac{2x-6}{2} =3}
We can solve this with the help of the balance method in the following:
\latex{\frac{2x-6}{2} =3} \latex{/\times2}
\latex{ 2x-6 = 6 \;\;/+6 }
\latex{ 2x = 12 \;\;/\div 2 }
\latex{ x = 6. }
So the number thought of is \latex{6;} we have already made sure of the correctness of it in the previous solution.
Example 2
Let us solve the following equation:
\latex{3\left(x-2\right) -2\left(-2x+3\right) +1=\frac{13x+7}{7} -\frac{12x}{14}}.
Solution
During the solution when doing the transformations our aim is that the equation becomes simpler step by step, and that in the end only the unknown stands on one side.
On the left side let us open the parentheses, and on the right side let us simplify the second fraction, thus the two fractions will have a common denominator.
\latex{3x-6+4x-6+1=\frac{13x+7-6x}{7}},
\latex{7x-11=\frac{7x+7}{7}}.
\latex{7x-11=\frac{7x+7}{7}}.
After the combinations and simplifications we arrange the unknowns to the left side of the equation, and we arrange the constant terms to the right side.
\latex{ 7x - 11 = x + 1, }
\latex{ 6x = 12, }
\latex{ x = 2. }
\latex{ 6x = 12, }
\latex{ x = 2. }
Based on the checking we can make sure that \latex{ x = 2 } is indeed a solution of the equation.
Example 3
Let us solve the following equation:
\latex{5x-\frac{3}{2x-10}=25-\frac{3}{2x-10} }.
Solution
The domain of the equation cannot contain the number for which \latex{ 2x - 10 = 0 }, since the fraction does not have a meaning then. It means that every real number can be found in the domain except for \latex{ 5 }.
We add \latex{\frac{3}{2x-10}} to both sides of the equation, and thus we get the following equation:
\latex{ 5x = 25, }
\latex{ x = 5. }
\latex{ x = 5. }
By adding the fraction we added an expression containing an unknown to both sides of the equation, and thus we changed its domain. The domain of the resulting equation became the set of the real numbers, and such a solution resulted which cannot be a root of the
original equation because of the given conditions. Therefore this result is called a false root. The equation does not have a solution.
original equation because of the given conditions. Therefore this result is called a false root. The equation does not have a solution.
Example 4
Let us solve the following.
x(3x – 8)(2x + 3) = 6x3 – 7x2 + 24.
Solution
The domain of the equation: \latex{\R}.
First let us expand the parentheses on the left side, and then let us combine the like terms.
\latex{ x(6x^2 + 9x - 16x - 24) = 6x^3 - 7x^2 + 24, }
\latex{ x(6x^2 - 7x - 24) = 6x^3 - 7x^2 + 24, }
\latex{ 6x^3 - 7x^2 - 24x = 6x^3 - 7x^2 + 24, }
\latex{ - 24x = 24, }\latex{ x = -1. }
\latex{ x(6x^2 - 7x - 24) = 6x^3 - 7x^2 + 24, }
\latex{ 6x^3 - 7x^2 - 24x = 6x^3 - 7x^2 + 24, }
\latex{ - 24x = 24, }\latex{ x = -1. }
It can be observed that we did not do any such step during the transformations with which we would have changed the domain of the equation or we would have got new roots. Every step was an equivalent transformation. By substitution we can make sure of the correctness of the resulting root.
Exercises
{{exercise_number}}. Solve the following equations on the set of real numbers.
- \latex{2\left(2x+1\right) -1=1-2\left(1+2x\right) }
- \latex{\frac{2y}{3} -\frac{3y}{2} =\frac{1}{6} }
- \latex{2\frac{1}{3}z-1\frac{3}{5}\left(5-z\right) =1}
- \latex{2\left[2\left(2v-1\right)-1 \right] -1=0}
{{exercise_number}}. Solve the following equations on the set of real numbers.
- \latex{2\left(x+1\right) \frac{1}{x} -1=1-\frac{2\left(1+2x\right) }{x} }
- \latex{\frac{2y}{y+1} -\frac{3y}{y+1} =\frac{1}{6} }
- \latex{\frac{z}{3z-3} -\frac{5-z}{z-1} =1}
- \latex{\frac{v-1}{v+1}-\frac{2v-1}{2v+1}=0 }
