The value of x is 1.
The value of y is 4.
Solution:
Given TQRS is a rhombus.
<u>Property of rhombus:
</u>
Diagonals bisect each other.
In diagonal TR
⇒ 3x + 2 = y + 1
⇒ 3x – y = –1 – – – – (1)
In diagonal QS
⇒ x + 3 = y
⇒ x – y = –3 – – – – (2)
Solve (1) and (2) by subtracting
⇒ 3x – y – (x – y) = –1 – (–3)
⇒ 3x – y – x + y = –1 + 3
⇒ 2x = 2
⇒ x = 1
Substitute x = 1 in equation (2), we get
⇒ 1 – y = –3
⇒ –y = –3 – 1
⇒ –y = –4
⇒ y = 4
The value of x is 1.
The value of y is 4.
3x - 2y = 10
10x + 2y = 8 (everything *2)
minus
Therefore final equation is -7x = 2 B
Answer:
Step-by-step explanation:
first multiply 4 X 2= 8
8+2x=42
Now you are going to pass 8 to the other side by adding -
2x=42-8
2x=34
Now you are going to divide 34 between 2 to eliminate the 2 multiplying in the other side. 34 between 2= 17
X=17
Hope this helps :)
Answer:
(X+3)(X+8)
Step-by-step explanation:
In the attached file
Answer:
3.6
Step-by-step explanation:
