How do you solve $\dfrac{2}{3}x + 1 = \dfrac{1}{2}x$ ?
Answer
626.7k+ views
Hint: In this question, we are given an equation in terms of $x$ and we have been asked to find the value of $x$. Shift the terms such that the terms containing $x$ are on one side and the constants are on the other side. Then, take LCM of denominators and make the denominators equal. Subtract and shift the terms to get the value of $x$.
Complete step by step answer:
We are given an equation in terms of $x$, and we have to find the value of $x$ .
$\dfrac{2}{3}x + 1 = \dfrac{1}{2}x$ …. (given)
Shifting the terms containing variable on the right-hand side and leaving the constant on the left-hand side
$1 = \dfrac{1}{2}x - \dfrac{2}{3}x$
Next step involves taking the LCM on the right-hand side and making the denominator the same.
LCM of $2$ and $3$ $ = 6$
Making denominator equal,
$1 = \dfrac{{1x \times 3}}{{2 \times 3}} - \dfrac{{2x \times 2}}{{3 \times 2}}$
On simplifying, we get,
$1 = \dfrac{{3x}}{6} - \dfrac{{4x}}{6}$
Simplifying further,
$1 = \dfrac{{ - x}}{6}$
Shifting the denominator to the other side,
$1 \times 6 = - x$
$ \Rightarrow - 6 = x$
Hence, after solving, we know that $x = - 6$.
Note: The given equation is a linear equation in one variable. An equation is a combination of variables and constants. When the highest degree of these variables is one, then the equation is called a linear equation. On the other hand, when the highest degree is two, it is called a quadratic equation.
A fact about these equations and degrees is that the number of solutions of an equation is equal to the degree of the equation. In other words, if an equation has a degree 1, then you will get only one value of $x$. If an equation has degree 2, then you will get two different or same values of $x$. This rule applies for every degree and values of $x$.
Complete step by step answer:
We are given an equation in terms of $x$, and we have to find the value of $x$ .
$\dfrac{2}{3}x + 1 = \dfrac{1}{2}x$ …. (given)
Shifting the terms containing variable on the right-hand side and leaving the constant on the left-hand side
$1 = \dfrac{1}{2}x - \dfrac{2}{3}x$
Next step involves taking the LCM on the right-hand side and making the denominator the same.
LCM of $2$ and $3$ $ = 6$
Making denominator equal,
$1 = \dfrac{{1x \times 3}}{{2 \times 3}} - \dfrac{{2x \times 2}}{{3 \times 2}}$
On simplifying, we get,
$1 = \dfrac{{3x}}{6} - \dfrac{{4x}}{6}$
Simplifying further,
$1 = \dfrac{{ - x}}{6}$
Shifting the denominator to the other side,
$1 \times 6 = - x$
$ \Rightarrow - 6 = x$
Hence, after solving, we know that $x = - 6$.
Note: The given equation is a linear equation in one variable. An equation is a combination of variables and constants. When the highest degree of these variables is one, then the equation is called a linear equation. On the other hand, when the highest degree is two, it is called a quadratic equation.
A fact about these equations and degrees is that the number of solutions of an equation is equal to the degree of the equation. In other words, if an equation has a degree 1, then you will get only one value of $x$. If an equation has degree 2, then you will get two different or same values of $x$. This rule applies for every degree and values of $x$.
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