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Hint- Here, we are given with one equation in one variable (x) which can be easily solved by using the basic principles of algebra. Here, we will rearrange the given equation so that we can get the value for variable x.

Complete step-by-step answer:

The given equation is $

5\left( {8x + 3} \right) = 9\left( {4x + 7} \right) \\

\Rightarrow 40x + 15 = 36x + 63{\text{ }} \to {\text{(1)}} \\

$

Now, the given equation is reduced to equation (1) from which we can easily say that in this equation (1) there is only one variable x.

Shifting the term 36x (present on the RHS) towards LHS and the term 15 (present on the LHS) towards RHS in equation (1), we get

$

\Rightarrow 40x - 36x = 63 - 15 \\

\Rightarrow 4x = 48 \\

$

Now, by shifting 4 (which is getting multiplied on the LHS) to the RHS it will be divided by 48 in the above equation.

$

\Rightarrow 4x = 48 \\

\Rightarrow x = \dfrac{{48}}{4} = 12 \\

$

So, the required value of x which satisfies the given equation is 12.

Note- In this particular problem, we have used the concepts like when a term is shifted from the LHS to the RHS or from the RHS to the LHS of any equation, the sign of that term changes. Also, when a number which is multiplied on the LHS or the RHS is shifted to the RHS or the LHS of that equation respectively, that number comes in division.

Complete step-by-step answer:

The given equation is $

5\left( {8x + 3} \right) = 9\left( {4x + 7} \right) \\

\Rightarrow 40x + 15 = 36x + 63{\text{ }} \to {\text{(1)}} \\

$

Now, the given equation is reduced to equation (1) from which we can easily say that in this equation (1) there is only one variable x.

Shifting the term 36x (present on the RHS) towards LHS and the term 15 (present on the LHS) towards RHS in equation (1), we get

$

\Rightarrow 40x - 36x = 63 - 15 \\

\Rightarrow 4x = 48 \\

$

Now, by shifting 4 (which is getting multiplied on the LHS) to the RHS it will be divided by 48 in the above equation.

$

\Rightarrow 4x = 48 \\

\Rightarrow x = \dfrac{{48}}{4} = 12 \\

$

So, the required value of x which satisfies the given equation is 12.

Note- In this particular problem, we have used the concepts like when a term is shifted from the LHS to the RHS or from the RHS to the LHS of any equation, the sign of that term changes. Also, when a number which is multiplied on the LHS or the RHS is shifted to the RHS or the LHS of that equation respectively, that number comes in division.