Solve the equations.
$1 \frac{3}{4} x+4=22$ | 2 |
$4-7 x-3=2-7 x-5$ | 5 |
We are given two separate equations involving the variable \(x\). Our goal is to solve for \(x\) in both instances.
For the first equation, we will isolate \(x\) by simplifying the coefficients and applying algebraic operations. For the second equation, we will combine like terms on both sides to determine if there is a valid value for \(x\), if the equation is always true, or if it has no solution.
First, we convert the mixed fraction \(1 \frac{3}{4}\) to an improper fraction: \(1 \frac{3}{4} = \frac{1 \times 4 + 3}{4} = \frac{7}{4}\).
The equation becomes:
\[\frac{7}{4} x + 4 = 22\]
Subtract 4 from both sides to isolate the term with \(x\):
\[\frac{7}{4} x = 22 - 4\]
\[\frac{7}{4} x = 18\]
Multiply both sides by \(\frac{4}{7}\) to solve for \(x\):
\[x = 18 \cdot \frac{4}{7}\]
\[x = \frac{72}{7}\]
Thus, the solution to the first equation is \(x = \frac{72}{7}\).
We start with the equation:
\[4 - 7x - 3 = 2 - 7x - 5\]
First, simplify the left side:
\[(4 - 3) - 7x = 1 - 7x\]
Next, simplify the right side:
\[(2 - 5) - 7x = -3 - 7x\]
So the equation simplifies to:
\[1 - 7x = -3 - 7x\]
Add \(7x\) to both sides to eliminate the variable:
\[1 = -3\]
This statement is a contradiction, as \(1\) can never equal \(-3\). Therefore, the second equation has no solution.
Linear equations are used in many real-world scenarios to model relationships. For example, if a company has a fixed cost for production (represented by the constant term) and a variable cost per unit (the coefficient of \(x\)), they use linear equations to determine the break-even point or total expenses. Solving these equations helps businesses and individuals make informed financial decisions, such as budgeting for a project or determining the minimum units to sell for a profit.