Drag each equation to the correct location on the table. Solve the equations for the given variable. Then place the equations in the table under the correct solution.
| x=3 | x ≠ 3 |
|---|---|
-14x = -42 -(3/5) + x = (12/5) x/4 = 6/8 -6 + x = -9 x/3 = 9 x - 5 = -2
Answer (2)
Solve the equation − 14 x = − 42 and get x = 3 .
Solve the equation − 5 3 + x = 5 12 and get x = 3 .
Solve the equation 4 x = 8 6 and get x = 3 .
Solve the equation − 6 + x = − 9 and get x = − 3 .
Solve the equation 3 x = 9 and get x = 27 .
Solve the equation x − 5 = − 2 and get x = 3 .
The equations with x = 3 are − 14 x = − 42 , − 5 3 + x = 5 12 , 4 x = 8 6 , x − 5 = − 2 .
The equations with x = 3 are − 6 + x = − 9 , 3 x = 9 .
The final answer is the classification of the equations: x = 3 : − 14 x = − 42 , − 5 3 + x = 5 12 , 4 x = 8 6 , x − 5 = − 2 ; x = 3 : − 6 + x = − 9 , 3 x = 9 .
Explanation
Analyzing the Equations We are given six equations to solve for x and classify based on whether their solution is x = 3 or x = 3 .
The equations are:
− 14 x = − 42
− 5 3 + x = 5 12
4 x = 8 6
− 6 + x = − 9
3 x = 9
x − 5 = − 2
Solving Each Equation Let's solve each equation for x :
− 14 x = − 42 . Dividing both sides by − 14 , we get x = − 14 − 42 = 3 .
− 5 3 + x = 5 12 . Adding 5 3 to both sides, we get x = 5 12 + 5 3 = 5 15 = 3 .
4 x = 8 6 . Multiplying both sides by 4 , we get x = 4 ⋅ 8 6 = 8 24 = 3 .
− 6 + x = − 9 . Adding 6 to both sides, we get x = − 9 + 6 = − 3 .
3 x = 9 . Multiplying both sides by 3 , we get x = 3 ⋅ 9 = 27 .
x − 5 = − 2 . Adding 5 to both sides, we get x = − 2 + 5 = 3 .
Classifying the Equations Now, let's classify each equation based on its solution:
Equations with x = 3 are: − 14 x = − 42 , − 5 3 + x = 5 12 , 4 x = 8 6 , and x − 5 = − 2 .
Equations with x = 3 are: − 6 + x = − 9 (since x = − 3 ) and 3 x = 9 (since x = 27 ).
Final Classification Therefore, the equations are classified as follows:
x = 3 : − 14 x = − 42 , − 5 3 + x = 5 12 , 4 x = 8 6 , x − 5 = − 2
x = 3 : − 6 + x = − 9 , 3 x = 9
Examples
Imagine you're baking cookies and need to adjust ingredient quantities based on how many cookies you want to make. If the original recipe for 12 cookies requires 3 cups of flour, you can set up an equation to find out how much flour you need for 24 cookies. This involves solving a simple linear equation, similar to the ones we solved, to scale the ingredients correctly. Understanding how to solve these equations helps in everyday tasks like cooking, managing budgets, or planning projects where resources need to be adjusted proportionally.
The equations classify into those with solutions of x=3 and those with x≠3. The equations yielding x=3 are -14x = -42, -3/5 + x = 12/5, x/4 = 6/8, x - 5 = -2, while -6 + x = -9 and x/3 = 9 yield other values. Thus, the classifications are complete based on whether x equals 3 or not.
;
