No solution of the system of equations y = -2x + 5 and -5y = 10x + 20 ⇒ 2nd answer
Step-by-step explanation:
Let us revise the types of solutions of a system of linear equations
- One solution
- No solution when the coefficients of x and y in the two equations are equal and the numerical terms are different
- Infinitely many solutions when the coefficients of x , y and the numerical terms are equal in the two equations
∵ y = -2x + 5
- Add 2x to both sides
∴ 2x + y = 5 ⇒ (1)
∵ -5y = 10x + 20
- Subtract 10x from both sides
∴ -10x - 5y = 20
- Divide both sides by -5
∴ 2x + y = -4 ⇒ (2)
∵ The coefficient of x in equation (1) is 2
∵ The coefficient of x in equation (2) is 2
∴ The coefficients of x in the two equations are equal
∵ The coefficient of y in equation (1) is 1
∵ The coefficient of y in equation (2) is 1
∴ The coefficients of y in the two equations are equal
∵ The numerical term in equation (1) is 5
∵ The numerical term in equation (2) is -4
∴ The numerical terms are different
From the 2nd rule above
∴ No solution of the system of equations
No solution of the system of equations y = -2x + 5 and -5y = 10x + 20
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Answer: 4
Step-by-step explanation:
18-14=4
Answer:
33+12t−21t^2
Step-by-step explanation:
(2t-7)²-(5t-4)²
Use binomial theorem (a−b)^2 = a^2−2ab+b^2 to expand (2t-7)².
4t^2−28t+49−(5t-4)²
Use binomial theorem (a−b)^2 = a^2−2ab+b^2 to expand (5t-4)².
4t^2−28t+49−(25t^2−40t+16)
To find the opposite of 25t^2
−40t+16, find the opposite of each term.
4t^2−28t+49−25t^2−40t+16
Combine 4t^2 and −25t^2 to get −21t^2.
−21t^2−28t+49+40t−16
Combine −28t and 40t to get 12t.
−21t^2+12t+49−16
Subtract 16 from 49 to get 33.
−21t^2+12t+33
Swap terms to the left side.
33+12t−21t^2
I hope this helped!
Answer:
y=1/4x
Step-by-step explanation: in image below
Don’t follow the link he or she gave you.