q + 12 - 2(q - 22) > 0 |use distributive property
q + 12 +(-2)(q) + (-2)(-22) > 0
q + 12 - 2q + 44 > 0 |combine like terms
(q - 2q) + (12 + 44) > 0
-q + 56 > 0 |subtract 56 from both sdies
-q > -56 |change the signs
<h3>q < 56</h3>
Answer:
{-infinity, infinity}
no solution
Step-by-step explanation:
|x| +15≥4
Subtract 15 from each side
|x| +15-15≥4-15
|x| ≥-11
This is always true so x can be and real number
{-infinity, infinity}
-2|5x+3| = 8
Divide by -2
-2/-2|5x+3| = 8/-2
|5x+3| = -4
This can never happen because absolute values are zero or greater
no solution
You would just do 4 x 7, which means there are 28 combinations.
Q1. The answer is x = 1, y = 1, z = 0
<span>(i) -2x+2y+3z=0
</span><span>(ii) -2x-y+z=-3
</span>(iii) <span>2x+3y+3z=5
</span><span>_________
Sum up the first and the third equation:
</span>(i) -2x+2y+3z=0
(iii) 2x+3y+3z=5
_________
5y + 6z = 5
Sum up the second and the third equation:
(ii) -2x-y+z=-3
(iii) 2x+3y+3z=5
_________
2y + 4z = 2
(iv) 5y + 6z = 5
(v) 2y + 4z = 2
________
Divide the fifth equation by 2
(iv) 5y + 6z = 5
(v) y + 2z = 1
________
Multiple the second equation by -3 and sum the equation
(iv) 5y + 6z = 5
(v) -3y - 6z = -3
________
2y = 2
y = 2/2 = 1
y + 2z = 1
1 + 2z = 1
2z = 1 - 1
2z = 0
z = 0
-2x-y+z=-3
-2x - 1 + 0 = -3
-2x = -3 + 1
-2x = -2
x = -2/-2 = 1
Q2. The answer is x = -37, y = -84, z = -35
<span>(i) x-y-z=-8
(ii) -4x+4y+5z=7
(iii) 2x+2z=4
______
</span>Divide the third equation by 2 and rewrite z in the term of x:
(iii) x+z=2
z = 2 - x
______
Substitute z from the third equation and express y in the term of x:
<span>x-y-(2-x)=-8
x - y - 2 + x = 8
2x - y = 10
y = 2x - 10
______
Substitute z from the third equation and y from the first equation into the second equation:
</span><span>-4x + 4y + 5z = 7
-4x + 4(2x - 10) + 5(2 - x) = 7
-4x + 8x - 40 + 10 - 5x = 7
-x -30 = 7
-x = 30 + 7
x = -37
y = 2x - 10 = 2*(-37) - 10 = -74 - 10 = -84
z = 2 - x = 2 - 37 = -35</span>
Answer:
Step-by-step explanation:
Both 32 and 40 are divisible by 8.
32÷8=4
40÷8=5
Thus, the simplest form is
