We are asked to solve two distinct problems. The first problem is to find the values of \(x\) that satisfy the absolute value equation \(-2|x - 4| + 7 = -3\). The second problem is to solve the linear equation \(v = 7x - 3w\) for \(x\). We will address each problem separately using standard algebraic manipulation techniques.
To solve \(-2|x - 4| + 7 = -3\), we first isolate the absolute value term.
Subtract \(7\) from both sides:
\[-2|x - 4| = -3 - 7\]
\[-2|x - 4| = -10\]
Next, divide both sides by \(-2\):
\[|x - 4| = \frac{-10}{-2}\]
\[|x - 4| = 5\]
Now, we consider the definition of the absolute value, which gives us two cases:
Case 1: \(x - 4 = 5\)
Adding \(4\) to both sides, we get:
\[x = 5 + 4 = 9\]
Case 2: \(x - 4 = -5\)
Adding \(4\) to both sides, we get:
\[x = -5 + 4 = -1\]
Thus, the solutions are \(x = 9\) and \(x = -1\).
To solve \(v = 7x - 3w\) for \(x\), we need to isolate \(x\) on one side of the equation.
Start with the given equation:
\[v = 7x - 3w\]
Add \(3w\) to both sides:
\[v + 3w = 7x\]
Finally, divide both sides by \(7\) to solve for \(x\):
\[x = \frac{v + 3w}{7}\]
This gives us the expression for \(x\) in terms of \(v\) and \(w\).
The ability to manipulate equations and solve for variables is foundational to many real-world applications. For instance, in physics, a formula like \(d = v_0 t + \frac{1}{2}at^2\) relates distance, velocity, time, and acceleration. Solving this for time or acceleration requires the same algebraic isolation techniques used here. Similarly, in economics, a budget constraint equation \(I = p_1x_1 + p_2x_2\) can be rearranged to find the quantity of one good \(x_1\) in terms of the price of both goods and the other quantity \(x_2\). Mastery of these operations ensures that you can rearrange any physical or economic model to solve for the specific variable you need.