Solve the linear equation
$\dfrac{{y - 1}}{3} - \dfrac{{y - 2}}{4} = 1$
Answer
519.6k+ views
Hint: Linear equations can be defined as the equations with unknown variables present in it, with the degree of 1. Linear equations can be also regarded as the first order equations. If the above example is considered, that is a linear equation with just one variable, the steps are very simple and there can be different ways in which the question can be solved. Basic methods of addition, subtraction and multiplication are only used. We get the value of y by bringing all the terms of y to one side and keeping the rest on the other.
Complete step-by-step solution:
$\dfrac{{y - 1}}{3} - \dfrac{{y - 2}}{4} = 1$
To find: y
We will start solving the sum by solving the left hand side First.
LHS: $\dfrac{{y - 1}}{3} - \dfrac{{y - 2}}{4}$
We need to make the base the same in order to perform subtraction, so we have to take LCM.
On Taking LCM, we get $\dfrac{{4 \times (y - 1)}}{{4 \times 3}} - \dfrac{{3 \times (y - 2)}}{{3 \times 4}}$
Therefore, the LHS becomes, $\dfrac{{4y - 4 - 3y + 6}}{{12}}$
After replacing the LHS in the original equation, the equation becomes,
$\dfrac{{4y - 4 - 3y + 6}}{{12}} = 1$
On cross multiplying, we get¸ $4y - 4 - 3y + 6 = 12$
Further solving it, we get, $y + 2 = 12$
We want the value of y, so we keep y on one side taking all the other terms on other side, which results into:
\[y = 12 - 2 = 10\]
Therefore, $y = 10$
Note: Few things should be kept in mind when we come across such questions. Firstly look if the base of the fractions is the same in order to perform addition or subtraction. Secondly, addition, subtraction, all these methods are to be performed correctly, most of the mistakes happen because of the wrong sign. Just to be sure about your answer, you can put the value of y in the original equation and compare.
Complete step-by-step solution:
$\dfrac{{y - 1}}{3} - \dfrac{{y - 2}}{4} = 1$
To find: y
We will start solving the sum by solving the left hand side First.
LHS: $\dfrac{{y - 1}}{3} - \dfrac{{y - 2}}{4}$
We need to make the base the same in order to perform subtraction, so we have to take LCM.
On Taking LCM, we get $\dfrac{{4 \times (y - 1)}}{{4 \times 3}} - \dfrac{{3 \times (y - 2)}}{{3 \times 4}}$
Therefore, the LHS becomes, $\dfrac{{4y - 4 - 3y + 6}}{{12}}$
After replacing the LHS in the original equation, the equation becomes,
$\dfrac{{4y - 4 - 3y + 6}}{{12}} = 1$
On cross multiplying, we get¸ $4y - 4 - 3y + 6 = 12$
Further solving it, we get, $y + 2 = 12$
We want the value of y, so we keep y on one side taking all the other terms on other side, which results into:
\[y = 12 - 2 = 10\]
Therefore, $y = 10$
Note: Few things should be kept in mind when we come across such questions. Firstly look if the base of the fractions is the same in order to perform addition or subtraction. Secondly, addition, subtraction, all these methods are to be performed correctly, most of the mistakes happen because of the wrong sign. Just to be sure about your answer, you can put the value of y in the original equation and compare.
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