Question

Solve the linear equation: $ \dfrac{{3t - 2}}{4} - \dfrac{{2t + 3}}{3} = \dfrac{2}{3} - t $ .

Answer 1

Hint: As we know that the above given equation $ 3a + 2b = c $ is a linear equation. An equation for a straight line is called a linear equation. The standard form of linear equations in two variables is $ Ax + By = C $ . When an equation is given in this form it’s also pretty easy to find both intercepts $ (x,y) $ . In this question we will first solve the Left hand side and then the right hand side and then we equate them and get the value.

Complete step-by-step answer:
As we know that the above given equation is a linear equation and to solve for $ t $ . So we need to isolate the term containing $ t $ on the left hand side i.e. to simplify
 $ \dfrac{{3t - 2}}{4} - \dfrac{{2t + 3}}{3} = \dfrac{2}{3} - t $ .
 Here we have the left hand side $ \dfrac{{3t - 2}}{4} - \dfrac{{2t + 3}}{3} $ . We will take the L.C.M and then solve it:
 $ \dfrac{{3(3t - 2) - 4(2t + 3)}}{{12}} = \dfrac{{3 \times 3t - 3 \times 2 - 4 \times 2t - 4 \times 3}}{{12}} $ .
On further simplifying we have $ \dfrac{{9t - 6 - 8t - 12}}{{12}} = \dfrac{{t - 18}}{{12}} $ .
We will now take the right hand side of the equation i.e. $ \dfrac{2}{3} - t $ .
Similarly by taking the LCM we have $ \dfrac{{2 - 3t}}{3} $ .
Now we will equate the both LHS and RHS, : $ \dfrac{{t - 18}}{{12}} = \dfrac{{2 - 3t}}{3} $ .
By cross multiplication we can write it as $ 3(t - 18) = 12(2 - 3t) $ .
We will now break the brackets and multiply :
 $ 3 \times t - 3 \times 18 = 12 \times 2 - 12 \times 3t \Rightarrow 3t - 54 = 24 - 36t $ .
By grouping the similar terms together we have:
$ 3t + 36t = 24 + 54 \Rightarrow 39t = 78 $ .
So it gives us the value $ t = \dfrac{{78}}{{39}} = 2 $ .
Hence the required answer is $ t = 2 $ .
So, the correct answer is “ $ t = 2 $ ”.

Note: We should keep in mind the positive and negative signs while calculating the value of any variable as it will change it’s slope and value. In the equation $ Ax + By = C $ , $ A $ and $ B $ are real numbers and $ C $ is a constant, it can be equal to zero $ (0) $ also. These types of equations are of first order. Linear equations are also first-degree equations as it has the highest exponent of variables as $ 1 $ . The slope intercept form of a linear equation is $ y = mx + c $ ,where $ m $ is the slope of the line and $ b $ in the equation is the y-intercept and $ x $ and $ y $ are the coordinates of x-axis and y-axis , respectively.