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Write two equations that represent this situation.
See solution.
Statement Solving the first equation forx, we get x=11-y. Now, we can substitute 11-y for x in the second equation, 2( 11-y)+4y=36. Statement By substitution,2(11-y)+4y=36. Furthermore, we can use the Distributive Property to remove the parentheses. This gives us 22-2y+4y=36. Statement By the Distributive Property, 22-2y+4y=36. Next, we can combine like terms in our equation. This means that we will subtract 2y from 4y. Then, our equation becomes 22+2y=36. Statement Combining like terms we get 22+2y=36. Now, by the Subtraction Property of Equality we can subtract 21 from both sides, 2y=14. By the Division Property of Equality, we can divide the equation by 2, which gives us y=7. Statement By the Subtraction Property of Equality and the Division Property of Equality, we solve the equation for y and get y=7. Finally, we can substitute y= 7 in the first equation, x=11-y. This gives us x=11- 7, and hence x=4. Therefore, there were 4 bikes and 7 skateboards at the park. Statement By substitution,x=11-7. Calculating the difference, we getx=4. Therefore, there were4 bikes and 7skateboards at the park.
Let x be the number of bikes and y the number of skateboards. There were a total of 11 bikes and skateboards, so x+y=11. Each bike has 2 wheels and each skateboard has 4 wheels. The bikes and skateboards had a total of 36 wheels. 2x+4y=36 Solving the first equation for x, we get x=11-y. By substitution, 2(11-y)+4y=36. By the Distributive Property, 22-2y+4y=36. Now let's combine like terms. 22+2y=36 By the Substitution Property of Equality and the Division Property of Equality, we solve the equation for y and get y=7. By substitution, x=11-7. Calculating the difference, we get x=4. Therefore, there were 4 bikes and 7 skateboards at the park.