Tell whether the ordered pair is a solution of the equation. Just substitute the given x and y to see if the equation "works". Write "solution" if it works and "not a solution" if it doesn't. 1) y = 4x + 2; (2, 10) 2) 2x + y = 5; (7, 5) 3) y = 6 – x; (–3, 3) 4) x + 8y = 2; (10, –1) 5) y = 6x + 7; (2, 21)

(1) Solution

(2) Not a solution

(3)  Not a solution

(4) Solution

(5)Not a solution

Step-by-step explanation:

(1)

Given equation is

y= 4x + 2

(2,10) is a solution of the above equation. Then the point  will be satisfy the above equation.

Putting x= 2 and y = 10

∴ 10 = (4×2)+2

⇒10= 8+2

⇒10=10

Therefore (2,10) is a solution of the equation.

(2)

Given equation is

2x+y=5

(7,5)is a solution of the above equation. Then the point  will be satisfy the above equation.

Putting x= 7 and y = 5

∴(2×7)+5=5

⇒14+5=5

⇒19≠5

Therefore (7,5) is not a solution of the equation.

(3)

Given equation is

y=6-x

(-3,3)is a solution of the above equation. Then the point will be satisfy the above equation.

Putting x= -3 and y = 3

∴3=6-(-3)

⇒3=6+3

⇒3≠9

Therefore (-3,3) is not a solution of the equation.

(4)

Given equation is

x+8y=2

(10,-1) is a solution of the above equation. Then the point  will be satisfy the above equation.

Putting x= 10 and y = -1

∴ 10+8.(-1) = 2

⇒10-8= 2

⇒2=2

Therefore (10,-1) is a solution of the equation.

(5)

Given equation is

y=6x+7

(2,21) is a solution of the above equation. Then the point  will be satisfy the above equation.

Putting x= 2 and y = 21

∴ 21=(6×2)+7

⇒21= 12+7

⇒21≠19

Therefore (2,21) is not a solution of the equation.

dont think so

step-by-step explanation: