Problem Type

Complicated Linear Equations

In this question, all the coefficients are fractions. Such equations need to have integer coefficients to solve.

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Explanation

All of the coefficients are not integers. In this case, we can multiply an equal number. Multiplying an equal number on both sides keeps the relation.

The equation for us to solve is the following.

[tex]\text{$\Longrightarrow \dfrac{4x+3}{9}+\dfrac{2x}{5}=\dfrac{3x+2}{15}$}[/tex]

We don't want the fractions. What least number makes them integers? The LCM or the least common multiple has all the factors of the denominator.

[tex]\;[/tex]

Solution

45 is the LCM of the denominators.

[tex]\text{$\Longrightarrow 45\times \dfrac{4x+3}{9}+45\times \dfrac{2x}{5}=45\times \dfrac{3x+2}{15}$}[/tex]

We can see them vanish.

[tex]\text{$\Longrightarrow 5\times(4x+3)+9\times 2x=3\times (3x+2)$}[/tex]

Let's use the distributive property.

[tex]\text{$\Longrightarrow 20x+15+18x=9x+6$}[/tex]

Now it is a simple equation to solve.

[tex]\text{$\Longrightarrow 20x+18x-9x=6-15$}[/tex]

[tex]\text{$\Longrightarrow 29x=-9$}[/tex]

[tex]\text{$\Longrightarrow \boxed{x=-\dfrac{9}{29}}$}[/tex]

[tex]\;[/tex]

[This is the required answer.]

Ask more if you have doubts.

Answer:

Question :-

[tex]\mapsto \sf \dfrac{4x + 3}{9} + \dfrac{2x}{5} =\: \dfrac{3x + 2}{15}\\[/tex]

Given :-

[tex]\mapsto \sf \dfrac{4x + 3}{9} + \dfrac{2x}{5} =\: \dfrac{3x + 2}{19}\\[/tex]

To Find :-

[tex]\mapsto[/tex] What is the value of x.

Solution :-

[tex]\implies \sf\bold{\dfrac{4x + 3}{9} + \dfrac{2x}{5} =\: \dfrac{3x + 2}{15}}\\[/tex]

[tex]\implies \sf \dfrac{5(4x + 3) + 9(2x)}{45} =\: \dfrac{3x + 2}{15}\\[/tex]

[tex]\implies \sf \dfrac{20x + 15 + 18x}{45} =\: \dfrac{3x + 2}{15}\\[/tex]

[tex]\implies \sf \dfrac{20x + 18x + 15}{45} =\: \dfrac{3x + 2}{15}\\[/tex]

[tex]\implies \sf \dfrac{38x + 15}{45} =\: \dfrac{3x + 2}{15}\\[/tex]

By doing cross multiplication we get,

[tex]\implies \sf 45(3x + 2) =\: 15(38x + 15)\\[/tex]

[tex]\implies \sf 135x + 90 =\: 570x + 225\\[/tex]

[tex]\implies \sf 135x - 570x =\: 225 - 90\\[/tex]

[tex]\implies \sf - 435x =\: 135\\[/tex]

[tex]\implies \sf x =\: - \dfrac{\cancel{135}}{\cancel{435}}\\[/tex]

[tex]\implies \sf\bold{x =\: - \dfrac{9}{29}}\\[/tex]

[tex]\sf\boxed{\bold{\therefore\: The\: value\: of\: x\: is\: - \dfrac{9}{29}\: .}}\\[/tex]