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:
RT = 20
Step-by-step explanation:
Point S is on line segment
R------------S------------T
RS + ST = RT
Given
ST=3x-8
RT=4x
RS=4x-7,
Step 1
We find x
4x - 7 + 3x - 8 = 4x
4x + 3x -7 - 8 = 4x
7x - 15 = 4x
7x - 4x = 15
3x = 15
x = 15/3
x = 5
Step 2
We find RT
RT = 4x
x = 5
RT = 4 × 5
RT = 20
The numerical length of RT is 20
Answer:
Step-by-step explanation:
90/15=6/z
6=6z
z=1
If Teresa drove 936 miles in 13 hours, this would mean she would have driven 72 miles every hour. 72 times 9 equals 648. After 9 hours, she would have driven 648 miles.
The explicit formula for arithmetic sequence is:
an=a+(n-1)d
where:
a=first term
d=common difference
given:
a3=22
a(17)=-20
substituting this in our equation we get:
22=a+(3-1)d
22=a+2d
a=22-2d........i
also:
-20=a+(17-1)d
-20=a+16d.....ii
but substituting i in ii we get:
-20=22-2d+16d
-20-22=14d
-42=14d
d=-3
but:
a=22-2d
a=22-2(-3)
a=28
thus the formula will be:
an=28-3(n-1)
thus the first term will be 28
the 2nd term will be:
a2=28-3(2-1)
a2=25
the 3rd term will be:
a3=28-3(3-1)
a3=28-6
a3=22
a4=28-3(4-1)
a4=28-9
a4=15
a5=28-3(5-1)
a5=28-3(4)
a5=28-12
a5=15
