Solving Problems With Linear Equations

Solving Problems With Linear Equations-67
For example, the equation \(x = x 1\) (which means a number equals the consecutive number) does not have a solution, because this is never true.Actually, this equation is reduced to 1 = 0, which is impossible.First we remove the parentheses: the one on the right has a negative sign, so it changes the sign of all the elements inside it; the one on the right has a 3 multiplying it, so it multiplies every element inside it.

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This fact assures us that if there is a solution, there can only be one solution (except in special cases where there are infinite solutions).

We say if there is a solution because sometimes equations don not any solution.

We are going to multiply the equation by the least common multiple of the denominators (5, 2 and 3), which is 30: We only have one parenthesis that is multiplied by 15.

To remove it, we multiply the content by 15: Therefore the solution is \( x = 23/30 \).

In this equation, we have a nested parenthesis (parenthesis within a parenthesis) and it is multiplied by fractions.

But before we worry about this, we will multiply all the equation by the least common multiple of the denominators, 6: Now we are going to the parentheses: on the left there are two, but we will treat it like one.

We can see this by going to the linear equations problems section.

These equations are known as linear, because the monomial literal part does not have an exponent (for example, \(3x\) can be part of a linear equation, but \(3x^2\) cannot because it is quadratic), so represented in a chart appears as a straight line.

This means that whatever the value of x, the equations is always going to be true.

To be solved, a word problem must be translated into the language of mathematics, where we use symbols for numbers - known or unknown, and for mathematical operations.


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