<span>1. Solve the system by substitution -2x+y=-11, 3x-4y=11.
</span>-2x+y=-11<span>
y = 2x-11
</span><span>3x-4y=11
</span>3x-4(2x-11)=11
3x -8x + 44 = 11
-5x = -33
<span>x =33/5
y = 11/5
</span><span>2. Solve the system using elimination- 2x+6y=-12, 5x-5y=10.
</span><span>(-2x+6y=-12)5
</span>-10x + 30y = -60
<span>(5x-5y=10)2
</span>10x - 10y = 20
Adding the two equations,
<span>-10x + 30y = -60
</span><span>10x - 10y = 20
</span>
20y = -40
y = -2
x = 0
<span>3. What is the solution of the following system?-3x-2y=-12, 9x+6y=-9.
</span><span>-3x-2y=-12
</span>y = -3/2 x + 6
<span>9x+6y=-9
</span>y = -3/2 x - 3/2
Since the slopes are equal then they are parallel lines so they never meet at one point.
F(x)= 3.99 + .99x
X is the amount of days it is late
Answer:
It can either be rational or irrational
Step-by-step explanation:
Given the quadratic equation ax²+bx+5 = 0, to get the zeros of the function, we will factorize the function to have;
ax²+bx = -5
x(ax+b) = -5
x = -5 and ax+b = -5
From the second function we have;
ax+b = -5
ax = -5-b
x = -5-b/a
This shows that the other value of x can be rational or irrational depending on the values of a and b
