Drag the tiles to the correct boxes to complete the pairs. Not all tiles will be used.
Match each equation with the correct solution.
[tex]x=9[/tex]
[tex]x=\left\{-\frac{7}{3}\right.[/tex]
[tex]x:=-15[/tex]
[tex]x=15[/tex]
[tex]x=-8[/tex]
[tex]x=\frac{7}{3}[/tex]
[tex]x=-9[/tex]
[tex]\frac{1}{3}(5 x-9)=2\left(\frac{1}{3} x+6\right)[/tex]
[tex]5(x+7)-3(x-4)=7 x+2[/tex]
[tex]4(3 x+5)-3=9 x-7[/tex]
Answer (1)
Solve the first equation 3 1 ( 5 x − 9 ) = 2 ( 3 1 x + 6 ) and find x = 15 .
Solve the second equation 5 ( x + 7 ) − 3 ( x − 4 ) = 7 x + 2 and find x = 9 .
Solve the third equation 4 ( 3 x + 5 ) − 3 = 9 x − 7 and find x = − 8 .
Match the equations with their solutions: x = 15 , x = 9 , x = − 8 .
Explanation
Problem Analysis We are given three equations and a set of possible solutions. Our goal is to match each equation with its correct solution.
Solving Equation 1 Let's solve the first equation: 3 1 ( 5 x − 9 ) = 2 ( 3 1 x + 6 ) .
First, distribute the constants on both sides: 3 5 x − 3 = 3 2 x + 12 .
Next, subtract 3 2 x from both sides: 3 5 x − 3 2 x − 3 = 12 , which simplifies to x − 3 = 12 .
Finally, add 3 to both sides: x = 15 .
Solving Equation 2 Now, let's solve the second equation: 5 ( x + 7 ) − 3 ( x − 4 ) = 7 x + 2 .
First, distribute the constants on both sides: 5 x + 35 − 3 x + 12 = 7 x + 2 .
Combine like terms on the left side: 2 x + 47 = 7 x + 2 .
Subtract 2 x from both sides: 47 = 5 x + 2 .
Subtract 2 from both sides: 45 = 5 x .
Finally, divide by 5: x = 9 .
Solving Equation 3 Let's solve the third equation: 4 ( 3 x + 5 ) − 3 = 9 x − 7 .
First, distribute the constant on the left side: 12 x + 20 − 3 = 9 x − 7 .
Combine like terms on the left side: 12 x + 17 = 9 x − 7 .
Subtract 9 x from both sides: 3 x + 17 = − 7 .
Subtract 17 from both sides: 3 x = − 24 .
Finally, divide by 3: x = − 8 .
Matching Solutions We have found the solutions for each equation: Equation 1: x = 15 Equation 2: x = 9 Equation 3: x = − 8
Now, we match these solutions with the given options.
Final Answer The solutions are: 3 1 ( 5 x − 9 ) = 2 ( 3 1 x + 6 ) -> x = 15 5 ( x + 7 ) − 3 ( x − 4 ) = 7 x + 2 -> x = 9 4 ( 3 x + 5 ) − 3 = 9 x − 7 -> x = − 8
Examples
When balancing chemical equations, you often need to solve linear equations to find the stoichiometric coefficients. Similarly, in electrical circuits, you might use Kirchhoff's laws, which involve solving systems of linear equations to determine the currents and voltages in the circuit. These equations help ensure that the number of atoms is conserved in a chemical reaction or that the energy is conserved in an electrical circuit. Understanding how to solve these equations is crucial for accurate calculations and predictions in these fields.
