Part 1)
we have
------> equation A
------> equation B
Multiply by
the equation A
------> equation C
Multiply by
the equation B

-------> equation D
Adds equation C and equation D

therefore
<u>the answer Part 1) is the option A </u>

Part 2)
we have
------> equation A

Simplify Divide by
both sides

------> equation B
the lines A and B are parallel lines, because the slope m is equal
so
The system has no solution
therefore
<u>the answer Part 2) is the option D</u>
There is no x value as there is no solution to the system.
Part 3)
we have
------> equation A

------> equation B
substitute equation B in equation A
![4x+2[x-3]=6](https://tex.z-dn.net/?f=4x%2B2%5Bx-3%5D%3D6)



therefore
<u>the answer part 3) is the option D</u>

Part 4)
Let
x---------> The number of one-step equations
y---------> The number of two-step equations
we know that

-------> equation A
------> equation B
substitute equation A in equation B
![3[1,120-y]-2y=1,300](https://tex.z-dn.net/?f=3%5B1%2C120-y%5D-2y%3D1%2C300)




therefore
<u>the answer part 4) is the option D</u>

Answer:
7 1/2 square meters
Step-by-step explanation:
The area of a rectangle is the product of its length and width.
A = LW
A = (9 m)(5/6 m) = 45/6 m² = 7 1/2 m²
The area of Dina's new rug is 7 1/2 square meters.
Answer:
Interest = $75
Total to pay back = $375
Step-by-step explanation:
The total amount she will have to pay back is found by the formula:
= Amount borrowed * (1 + rate * years)
= 300 * ( 1 + 5% * 5)
= $375
The interest is therefore:
= Amount to be paid back - Amount borrowed
= 375 - 300
= $75
Answer:
43 minutes
Step-by-step explanation:
- <em>Let cycling speed be x and driving speed be y</em>
<u>Equations as per given, considering same distance in both cases:</u>
- 19x + 8y = 13x + 10y
- 19x - 13x = 10y - 8y
- 6x = 2y
- 3x = y
We see that driving speed is 3 times greater than cycling speed
Then 8 minutes driving = 8*3= 24 minutes of cycling, or 10 minutes of driving = 10*3= 30 minutes of cycling:
- 19 + 24 = 43 minutes or
- 13 + 10*3 = 43 minutes is the time to cycle between A and B