$2(-6 m-3)=6(5 m-1)$
What is the solution to the equation?
The given equation is \(2(-6m - 3) = 6(5m - 1)\). This is a linear equation in one variable, \(m\). Our objective is to determine which of the provided options corresponds to the solution: no solution, \(m=0\), \(m=5\), or \(m=-3\). To solve this, we will simplify both sides by distributing the coefficients and then isolate the variable \(m\).
First, we distribute the constants on both sides of the equation.
On the left side:
\[2(-6m - 3) = 2 \times (-6m) + 2 \times (-3) = -12m - 6\]
On the right side:
\[6(5m - 1) = 6 \times 5m + 6 \times (-1) = 30m - 6\]
Now, we equate the simplified expressions:
\[-12m - 6 = 30m - 6\]
Next, we isolate the variable \(m\). We can start by adding 6 to both sides of the equation:
\[-12m - 6 + 6 = 30m - 6 + 6\]
\[-12m = 30m\]
Now, subtract \(-12m\) from both sides (or add \(12m\) to both sides) to gather the terms involving \(m\):
\[0 = 30m + 12m\]
\[0 = 42m\]
Dividing both sides by 42:
\[m = \frac{0}{42} = 0\]
Finally, we verify the solution by substituting \(m = 0\) back into the original equation:
Left side: \(2(-6(0) - 3) = 2(0 - 3) = 2(-3) = -6\)
Right side: \(6(5(0) - 1) = 6(0 - 1) = 6(-1) = -6\)
Since both sides are equal, the solution is indeed \(m = 0\). This matches one of the provided options.
Linear equations are used in many everyday situations involving balancing two different scenarios. For example, imagine two different cell phone plans. Plan A costs a base fee of \(\$6\) with a \(\$12\) charge for every hour used, while Plan B has a \(\$6\) base fee with a \(\$30\) charge for every hour. By setting up an equation like this, you can determine how many hours you need to use your phone so that the costs are equal, allowing you to choose the most cost-effective option. In this specific case, the equation confirms that if usage is zero, both plans cost the same amount (the base fee).