Hi there!
Here are some examples for you:
Two one step equations:
6x = 36
y + 5 = 10
Two equations containing fractions:
1/2x = 4
3/4y + 1 = 8
One equation using the distributive property:
2(x + 5) = 3
One equation with a decimal:
2.5x + 5 = 10
One real world problem that is solved by an equation:
At the beginning, I had 5 cups of flour. I made a batch of cookies. Now, I have 3.5 cups of flour. How much flour did I use?
5 - x = 3.5
Hope this helps!! :)
If there's anything else that I can help you with, please let me know!
Answer:
The larger number is 37. This makes the smaller number 56 - 37, which is 19.
Step-by-step explanation:
The first step is to pick variables for the two numbers. Lets call the smaller number x and the larger number y.
Since the sum of the two numbers is 56, this means that x + y = 56. Let's call this equation 1 and save it for later.
The next sentence says:
The smaller number is 18 less than the larger number. So, 18 less than the larger number would be y - 18. So, the smaller number must be equal to this, so
x = y - 18
Replace the x in equation 1 with y - 18
x + y = 56
(y-18) + y = 56 (replace x with y-18)
2y - 18 = 56 (add the y's)
2y = 74 (add 18 to both sides)
y = 37 (divide both sides by 2)
So, the larger number is 37. This makes the smaller number 56 - 37, which is 19.
Hope this helps!! :)
Answer:
There are 60 questions on the test
Step-by-step explanation:
Hope this helps :)
![q(x) = {x}^{2} + 3x + 4](https://tex.z-dn.net/?f=q%28x%29%20%3D%20%20%7Bx%7D%5E%7B2%7D%20%2B%203x%20%2B%204%20)
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Step(1)
To find q(a) we just need to put a instead of x in q(x) function.
Let's do it...
![q(a) = {a}^{2} + 3a + 4](https://tex.z-dn.net/?f=q%28a%29%20%3D%20%20%7Ba%7D%5E%7B2%7D%20%2B%203a%20%2B%204%20)
Multiply sides by -2 :
![- 2q(a) = - 2( {a}^{2} + 3a + 4)](https://tex.z-dn.net/?f=%20-%202q%28a%29%20%3D%20%20-%202%28%20%7Ba%7D%5E%7B2%7D%20%2B%203a%20%2B%204%29%20)
![- 2q(a) = - 2 {a}^{2} - 6a - 8](https://tex.z-dn.net/?f=%20-%202q%28a%29%20%3D%20%20-%202%20%7Ba%7D%5E%7B2%7D%20-%206a%20-%208%20)
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Step (2)
To find q(a+1) we just need to put a+1 instead of x in q(x) function.
Let's do it...
![q(a + 1) = ({a + 1})^{2} + 3(a + 1) + 4 \\](https://tex.z-dn.net/?f=q%28a%20%2B%201%29%20%3D%20%20%28%7Ba%20%2B%201%7D%29%5E%7B2%7D%20%2B%203%28a%20%2B%201%29%20%2B%204%20%5C%5C%20%20)
![q(a + 1) = {a}^{2} + 2a + 1 + 3a + 3 + 4 \\](https://tex.z-dn.net/?f=q%28a%20%2B%201%29%20%3D%20%20%7Ba%7D%5E%7B2%7D%20%2B%202a%20%2B%201%20%2B%203a%20%2B%203%20%2B%204%20%5C%5C%20%20)
![q(a + 1) = {a}^{2} + 5a + 8](https://tex.z-dn.net/?f=q%28a%20%2B%201%29%20%3D%20%20%7Ba%7D%5E%7B2%7D%20%2B%205a%20%2B%208%20)
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Step (3)
![q(a + 1) - 2q(a) =](https://tex.z-dn.net/?f=q%28a%20%2B%201%29%20-%202q%28a%29%20%3D%20)
![{a}^{2} + 5a + 8 - 2 {a}^{2} - 6a - 8 = \\](https://tex.z-dn.net/?f=%20%7Ba%7D%5E%7B2%7D%20%2B%205a%20%2B%208%20-%202%20%7Ba%7D%5E%7B2%7D%20-%206a%20-%208%20%3D%20%20%5C%5C%20%20%20)
![- {a}^{2} - a = - a(a + 1)](https://tex.z-dn.net/?f=%20-%20%20%7Ba%7D%5E%7B2%7D%20-%20a%20%20%3D%20%20-%20a%28a%20%2B%201%29)
And we're done.
Thanks for watching buddy good luck.
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