Question

Algebraically solve the following system of equations.
(x - 2)2 + (y - 3)2 = 16
x + y - 1 = 0

Answer 1

Answer:

um

Step-by-step explanation:

if it split into 2 then for the first one its x= -y + 13

for the second one its x= -y + 1

Answer 2

The solutions to the system of equations are (2, -1) and (-2,3)

How to solve the system of equations?

The equations are given as:

$(x - 2)^2 + (y - 3)^2 = 16$

x  + y - 1 = 0

Make y the subject in x  + y - 1 = 0

y = 1 - x

Substitute y = 1 - x in $(x - 2)^2 + (y - 3)^2 = 16$

$(x - 2)^2 + (1 - x - 3)^2 = 16$

Evaluate

$(x - 2)^2 + (-x - 2)^2 = 16$

Expand the equations

$x^2 - 4x + 4 + x^2 + 4x + 4 = 16$

Evaluate the like terms

$2x^2 + 8 = 16$

Subtract 8 from both sides

$2x^2 = 8$

Divide by 2

$x^2 = 4$

Take the square roots of both sides

x = ±2

Substitute x = ±2 in y = 1 - x

y = 1 - 2 = -1

y = 1 + 2 = 3

Hence, the solutions to the system of equations are (2, -1) and (-2,3)

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