2y-x=5
x2+y2-25=0

solve the system by the substitution method

(- 5, 0), (3, 4)

Step-by-step explanation:

Given the 2 equations

2y - x = 5 → (1)

x² + y² - 25 = 0 → (2)

Rearrange (1) expressing x in terms of y

x = 2y - 5 → (3)

Substitute x = 2y - 5 into (2)

(2y - 5)² + y² - 25 = 0 ← expand parenthesis and simplify

4y² - 20y + 25 + y² - 25 = 0

5y² - 20y = 0 ← factor out 5y from each term

5y(y - 4) = 0

Equate each factor to zero and solve for y

5y = 0 ⇒ y = 0

y - 4 = 0 ⇒ y = 4

Substitute these values into (3) for corresponding values of x

y = 0 : x = 0 - 5 = - 5 ⇒ (- 5, 0)

y = 4 : x = 8 - 5 = 3 ⇒ (3, 4)

y=0 or 4, x=-5 or 3

Step-by-step explanation:

First make one unknown in the first equation the subject of the formula.

2y-x=5 will become:x=2y-5

Then in the second equation, wherever there is x we substitute it with 2y-5

(2y-5)²+y²-25=0         (2y-5)(2y-5)+y²-25=0

opening the brackets gives: 4y²-10y-10y+25+y²-25=0

5y²-20y=0

Factorizing the equation we get the following:

y(5y-20)=0

y=0 or

5y-20=0

5y=20

y=4

Thus y=0 or 4

Substitute 0 and 4 for y in any of the equations provided.

Picking the first equation and starting with y=0

2(0)-x=5

x=-5

Using y=4

2(4)-x=5

x=3

δabc ≅ δdcb by sas because

db ≅ ac . . given (side)

bc ≅ bc . . reflexive property (side)

∠dbc ≅ ∠ acb . . alternate interior angles where transversal bc crosses parallel lines bd and ac (included angle)