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How do you solve $3\left( 3a-2b \right)=2\left( a-b \right)$ ?

Answer
VerifiedVerified
538.2k+ views
Hint: This problem contains two variables and we are given only one equation. So we need to express one variable in terms of the other one. For this, we first open the bracket on both the sides and multiply the terms. Then, we bring the like terms to one side and finally equate them by simplifying.

Complete step by step solution:
The given equation that we are provided is,
$3\left( 3a-2b \right)=2\left( a-b \right)$
We start off the solution by applying the distributive property to the first term of the left hand side of the equation. The distributive property states that an expression of the form \[x\left( y+z \right)\] can be written as $xy+yz$ . By comparing the first term of the above equation with the general form, we get,
$x=3,y=3a,z=-2b$
Thus, applying the distributive property, it becomes, $3\left( 3a-2b \right)=3\times 3a+3\times \left( -2b \right)$ which upon simplification gives, $9a-6b$ . Thus, the equation can be rewritten as,
$\Rightarrow 9a-6b=2\left( a-b \right)$
We then apply the distributive property to the first term of the right hand side of the equation. By comparing the first term of the right hand side of the above equation with the general form, we get,
$x=2,y=a,z=-b$
Thus, applying the distributive property, it becomes, $2\left( a-b \right)=2\times a+2\times \left( -b \right)$ which upon simplification give$\Rightarrow 9a-6b=2\left( a-b \right)$s, $2a-2b$ . Thus, the equation can be rewritten as,
$\Rightarrow 9a-6b=2a-2b$
The above rewritten equation has two variables $a$ and $b$ . So, we need to bring all the terms containing $a$ to the left hand side and all the terms containing $b$ to the right hand side. Doing so, the equation becomes,
$\Rightarrow 9a-2a=6b-2b$
This upon simplification gives,
$\Rightarrow 7a=4b$
Dividing both sides of the equation by $7$ gives,
$\Rightarrow a=\dfrac{4}{7}b$
Therefore, we can conclude that the solution of the equation is $a=\dfrac{4}{7}b$ .

Note: These problems only require simple multiplication and additions. Even then, mistakes are possible in considering the signs. So, we should be careful with that. Also, if we had been given another equation, then we would have to solve for both $a$ and $b$ , instead of expressing one in terms of other.

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