Question

Solve the following equations-
(A) $\dfrac{{4x + 7}}{{9 - 3x}} = \dfrac{1}{4}$
(B) $\dfrac{{3t - 2}}{4} - \dfrac{{2t + 3}}{4} = \dfrac{2}{3} - t$
(C) $\dfrac{{2 - 7x}}{{1 - 5x}} = \dfrac{{3 + 7x}}{{4 + 5x}}$
(D) $3\left( {5x - 7} \right) - 2\left( {9x - 11} \right) = 4\left( {8x - 13} \right) - 17$

Answer 1

Hint:
Since all the given equations are one degree and one variable, we can solve them easily by using algebraic operations on terms and transforming the equations. Fractional terms must be changed using the transposition of the denominator to avoid complications in the solution. Use the distributive property to expand the parenthesis, i.e. $a\left( {b + c} \right) = a \times b + a \times c$.

Complete step by step solution:
Here in this problem, we are given four equations with only one unknown variable each. And we need to find the values of the variables by solving each equation separately.
Let’s take each of the equations into the consideration separately to solve one by one.
For equation (A) we have: $\dfrac{{4x + 7}}{{9 - 3x}} = \dfrac{1}{4}$
To solve this, we can transpose the denominator of the left-side numerator and the denominator of the right-side to the numerator of the left-side fraction.
$ \Rightarrow 4 \times \left( {4x + 7} \right) = 1 \times \left( {9 - 3x} \right)$
Now we can expand the expressions on both sides by opening the parenthesis:
$ \Rightarrow 4 \times 4x + 4 \times 7 = 9 - 3x \Rightarrow 16x + 28 = 9 - 3x$
We can now take all the constants on one side and the term with the variable on the other:
$ \Rightarrow 16x + 28 = 9 - 3x \Rightarrow 16x + 3x = 9 - 28$
On further solving it, we get:
$ \Rightarrow 16x + 3x = 9 - 28 \Rightarrow 19x = - 19$
On dividing both sides by $19$ , we get:
$ \Rightarrow 19x = - 19 \Rightarrow x = - 1$
Hence, we solved the equation and obtained $x = - 1$
For equation (B) we have: $\dfrac{{3t - 2}}{4} - \dfrac{{2t + 3}}{4} = \dfrac{2}{3} - t$
Here we have an equation with variable ‘t’ in it. This can be simplified by first solving the left side fractions. Since they have the same denominator, we can write it as:
$ \Rightarrow \dfrac{{3t - 2}}{4} - \dfrac{{2t + 3}}{4} = \dfrac{2}{3} - t \Rightarrow \dfrac{{3t - 2 - \left( {2t + 3} \right)}}{4} = \dfrac{2}{3} - t$
After solving the numerator on the left-hand side of the equation, we get:
$ \Rightarrow \dfrac{{3t - 2 - \left( {2t + 3} \right)}}{4} = \dfrac{2}{3} - t \Rightarrow \dfrac{{3t - 2 - 2t - 3}}{4} = \dfrac{{t - 5}}{4} = \dfrac{2}{3} - t$
Now let’s separate the variable terms and the constants on different sides of the equation:
$ \Rightarrow \dfrac{{t - 5}}{4} = \dfrac{2}{3} - t \Rightarrow \dfrac{t}{4} - \dfrac{5}{4} = \dfrac{2}{3} - t \Rightarrow \dfrac{t}{4} + t = \dfrac{2}{3} + \dfrac{5}{4}$
After solving both sides, we have:
$ \Rightarrow \dfrac{t}{4} + t = \dfrac{2}{3} + \dfrac{5}{4} \Rightarrow \dfrac{{t + 4t}}{4} = \dfrac{{2 \times 4 + 5 \times 3}}{{3 \times 4}} = \dfrac{{8 + 15}}{{12}} \Rightarrow \dfrac{{5t}}{4} = \dfrac{{23}}{{12}}$
Now transposing the coefficient of ‘t’ to the right-hand side, we get:
$ \Rightarrow \dfrac{{5t}}{4} = \dfrac{{23}}{{12}} \Rightarrow t = \dfrac{{23 \times 4}}{{12 \times 5}} = \dfrac{{23}}{{15}}$
Hence, the equation (B) is solved and we get $t = \dfrac{{23}}{{15}}$
For equation (C), we have $\dfrac{{2 - 7x}}{{1 - 5x}} = \dfrac{{3 + 7x}}{{4 + 5x}}$
This can be solved by transposing the denominators of both sides to the opposite sides
$ \Rightarrow \dfrac{{2 - 7x}}{{1 - 5x}} = \dfrac{{3 + 7x}}{{4 + 5x}} \Rightarrow \left( {2 - 7x} \right)\left( {4 + 5x} \right) = \left( {3 + 7x} \right)\left( {1 - 5x} \right)$
Now we can expand the expressions by opening the parenthesis
$ \Rightarrow \left( {2 - 7x} \right)\left( {4 + 5x} \right) = \left( {3 + 7x} \right)\left( {1 - 5x} \right) \Rightarrow 2\left( {4 + 5x} \right) - 7x\left( {4 + 5x} \right) = 3\left( {1 - 5x} \right) + 7x\left( {1 - 5x} \right)$
On simplifying it further, we can see:
$ \Rightarrow 8 + 10x - 28x - 35{x^2} = 3 - 15x + 7x - 35{x^2} \Rightarrow 8 + 10x - 28x = 3 - 15x + 7x$
After separating the variable and constants on each side, we get:
$ \Rightarrow 8 + 10x - 28x = 3 - 15x + 7x \Rightarrow - 18x + 8x = - 8 + 3 \Rightarrow - 10x = - 5$
Multiplying both sides with $ - 10$ , we get $ \Rightarrow - 10x = - 5 \Rightarrow x = \dfrac{{ - 5}}{{ - 10}} = \dfrac{1}{2}$
Hence, we solved the equation and obtained the value $x = \dfrac{1}{2}$
For equation (D), we have $3\left( {5x - 7} \right) - 2\left( {9x - 11} \right) = 4\left( {8x - 13} \right) - 17$
This can be solved by opening up the parenthesis and expanding the expression:
$ \Rightarrow 3 \times 5x - 3 \times 7 - 2 \times 9x + 2 \times 11 = 4 \times 8x - 13 \times 4 - 17 \Rightarrow 15x - 21 - 18x + 22 = 32x - 52 - 17$
Now let’s separate the variables and constants on each side:
$ \Rightarrow 15x - 21 - 18x + 22 = 32x - 52 - 17 \Rightarrow 15x - 18x - 32x = 21 - 22 - 52 - 17 \Rightarrow {- 35x} = - 70$
On dividing both sides of the equation with $ - 35$ , we get:
$ \Rightarrow - 35x = - 70 \Rightarrow x = \dfrac{{ - 70}}{{ - 35}} = 2$

Hence, this equation is solved and we got the value $x = 2$.

Note:
In the above solution we expanded the parenthesis in multiple steps which was a use of the distributive property of addition and multiplication, i.e. $a\left( {m + n} \right) = a \times m + a \times n$ . And notice that the method addition and subtraction of fractions can differ according to the denominator of each of them. When the denominator of both the fraction is not similar, then we make them same using $\dfrac{a}{b} + \dfrac{c}{d} = \dfrac{{ad + cb}}{{bd}}$.