Answer: No
Solution:
Substitute the values of x & y into the first equation
6 -(-1) = 3
6 + 5 is NOT equal to 3
Therefore we can say this is not true
For this to be true, when substituting the values of x & y, the equation must be true for BOTH equations. Since it’s not true for the first equation it doesn’t matter if it’s true or not for the other equation.
I hope this helps!
Subtract 9a^2-6a+59a 2 −6a+59, a, squared, minus, 6, a, plus, 5 from 10a^2+3a+2510a 2 +3a+2510, a, squared, plus, 3, a, plus, 25
nasty-shy [4]
Answer:
The answer to your question is a² + 9a + 20
Step-by-step explanation:
10a² + 3a + 25 - (9a² - 6a + 5)
- Remove parentheses changing the sign of the second polynomial
10a² + 3a + 25 - 9a² + 6a - 5
- Group like terms
(10a² - 9a²) + (3a + 6a) + (25 - 5)
- Simplify and result
a² + 9a + 20
<u>answer:</u> 
<u>explanation:</u>
| distribute the -5.5 into the parentheses
| subtract 11 and move it over to -12.1
| divide by -5.5
| final answer
hope this helps! ❤ from peachimin
Finding the x-intercept first:
5 (x - 2) = y - 3
5x - 10 = 0 - 3
5x - 10 = -3
5x = 7
x = 7/5 or 1.4
Solve for y
y - 3 = 5 (0 - 2)
y - 3 = 0 - 10
y - 3 = -10
y = - 7
