<span>5ab^2 + 10ab
5ab^2 = 5ab *(b)
10ab = 5ab *(2)
so GCF of </span>5ab^2 + 10ab = 5ab
The formula for quadratic equation is given by:
x² - (sum)x + (product) = 0
so if we have real roots of 3 and 5
<span> x² - (3 + 5)x + (3*5) = 0
</span> x² - 8<span>x + 15 = 0
</span>
so if we have real roots of -3 and -6
<span> x² - (-3 - 6)x + (-2*-6) = 0
</span> x² - (-9<span>x) + 12 = 0
</span> x² + 9<span>x + 12 = 0</span>
Answer:
Solution: x = -2; y = 3 or (-2, 3)
Step-by-step explanation:
<u>Equation 1:</u> y = -5x - 7
<u>Equation 2:</u> -4x - 3y = -1
Substitute the value of y in Equation 1 into the Equation 2:
-4x - 3(-5x - 7) = -1
-4x +15x + 21 = -1
Combine like terms:
11x + 21 = - 1
Subtract 21 from both sides:
11x + 21 - 21 = - 1 - 21
11x = -22
Divide both sides by 11 to solve for x:
11x/11 = -22/11
x = -2
Now that we have the value for x, substitute x = 2 into Equation 2 to solve for y:
-4x - 3y = -1
-4(-2) - 3y = -1
8 - 3y = -1
Subtract 8 from both sides:
8 - 8 - 3y = -1 - 8
-3y = -9
Divide both sides by -3 to solve for y:
-3y/-3 = -9/-3
y = 3
Therefore, the solution to the given systems of linear equations is:
x = -2; y = 3 or (-2, 3)
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-8 because when you add a negative number your basically subtracting its absolute value.