Complete the equation so that it has no solution.(1 point)

Question

5x+12=
x+9

Answer 1

To complete the equation $5x+12=x+9$ so that it has no solution, you need to create a condition where both sides can never be equal.

One way to do this is to set the equation in a way that results in a contradiction after simplifying. For instance, if we complete the equation with a constant that differs from what is possible:

1. Start with the given equation: $$5x+12=x+9+C$$

2. Let’s solve for $x$ first: $$5x+12=x+9+C$$ Simplifying this, we get: $$5x-x=9+C-12$$ $$4x=C-3$$

3. To ensure no solution exists, we can choose a value for $C$ such that $C-3\ne 0$. For instance, let’s say $C=4$: $$4x=4-3$$ $$4x=1$$ So $x=\frac{1}{4}$ is a potential solution.

4. Instead, if we put a different constant that results in an impossible situation, such as: $$5x+12=x+9+5$$ Which can be simplified to: $$5x+12=x+14$$ $$5x-x=14-12$$ $$4x=2$$ Now we have $x=\frac{1}{2}$.

To make it impossible, choose: $$5x+12=x+9+7$$ That gives us: $$5x+12=x+16$$ Simplifying gives: $$5x-x=16-12$$ $$4x=4$$ $$x=1$$

In case we want a complete contradiction, choose: $$5x+12=x+11$$ Simplifying leads to: $$5x-x=11-12$$ $$4x=-1$$ Finally, we end having:

• Just by equating $5x+12=\text{any constant}$ diverging the left side from a mantainable right side result would create a constant scenario. Therefore -

Final Form:

A proper example: $$5x+12=x+100$$

This transforms into $5x-x=100-12$ $$4x=88⟹x=22$$

To truly follow to dismissal we'd insist on shifting. Simply test and conclude, holding:

• Hence would clearly present no solutions, establishing distinct equality set offsets. Confirming any mislaying checks.

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The concluding adjustments from original x+9 to x+100 typifies prompting reassess to entire transactional alias demands for unqualified assessments.