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Question: How do you solve \(6x-3=8x-9\)?...

How do you solve 6x−3=8x−96x-3=8x-9?

Explanation

Solution

The equation given in the above question, which is written as 6x−3=8x−96x-3=8x-9, is a linear equation in a single variable, which is x. Therefore, it will have a unique solution. Now, for solving the given equation we need to separate the variable terms on the LHS and the constant terms on the RHS. For this, we need to add 33 on both sides of the given equation to get 6x=8x−66x=8x-6. Then we have to subtract 8x8x from both sides to get −2x=−6-2x=-6. Finally, on dividing both the sides by −2-2, we will get the final solution of the given equation.

Complete step by step solution:
The equation given in the above question is
⇒6x−3=8x−9\Rightarrow 6x-3=8x-9
We can see that the highest power of the only variable x in the above equation is equal to one. Therefore, we can say that the given equation is a linear equation in a single variable. This means that it will have a unique solution.
Adding 33on both the sides of the above equation we get
⇒6x−3+3=8x−9+3 ⇒6x=8x−6 \begin{aligned} & \Rightarrow 6x-3+3=8x-9+3 \\\ & \Rightarrow 6x=8x-6 \\\ \end{aligned}
Now, we add 8x8x on both the sides of the above equation to get
⇒6x−8x=8x−6−8x ⇒−2x=−6 \begin{aligned} & \Rightarrow 6x-8x=8x-6-8x \\\ & \Rightarrow -2x=-6 \\\ \end{aligned}
Multiplying −1-1 both the sides, we get

& \Rightarrow -2x\left( -1 \right)=-6\left( -1 \right) \\\ & \Rightarrow 2x=6 \\\ \end{aligned}$$ Finally, we divide both the sides of the above equation by $$2$$ to get $\begin{aligned} & \Rightarrow \dfrac{2x}{2}=\dfrac{6}{2} \\\ & \Rightarrow x=3 \\\ \end{aligned}$ Hence, the final solution of the equation given in the above question is $x=3$. **Note:** Do not forget to back substitute the final obtained solution into the given equation and confirm whether the LHS is coming equal to RHS or not. We can also solve the equation using the graphical method by equating each side of the given equation to a variable y to get the equations $y=6x-3$ and $y=8x-9$. Then by considering the graphs of each of the equations, we can obtain the solution from the abscissa of the intersection point as below.