Question

Solve this equationX+y=3, x²+y²=29

Answer 1

Use the second equation to provide an expression for y in terms of x to substitute into the first equation to give a quadratic equation in x.

Explanation:

First add x to both sides of the second equation to get:

y=x+3

Then substitute this expression for y into the first equation to get:

29=x2+(x+3)2=2x2+6x+9

Subtract 29 from both ends to get:

0=2x2+6x−20

Divide both sides by 2 to get:

0=x2+3x−10=(x+5)(x−2)

So x=2 or x=−5

If x=2 then y=x+3=5.

If x=−5 then y=x+3=−2

So the two solutions (x,y) are (2,5) and (−5,−2)

Answer 2

x + y = 3 ___ [1] ( Given )


x² + y² = 29 ( Given )


$<b>( a + b )² = a² + b² + 2ab</b>$


Here, a = x, b = y


⇒ ( 3 )² = 29 + 2xy


⇒ 9 = 29 + 2xy


⇒ 9 - 29 = 2xy


⇒ - 20 = 2xy


⇒ xy = - 20 / 2


⇒ xy = - 10


⇒ x = - 10 / y


Put this value in [1], we get


⇒ ( - 10 / y ) + y = 3


⇒ ( - 10 + y² ) / y = 3


⇒ y² - 3y - 10 = 0


By Middle Term Factorisation


⇒ y² - 5y + 2y - 10 = 0


⇒ y ( y - 5 ) + 2 ( y - 5 ) = 0


⇒ ( y + 2 ) ( y - 5 ) = 0


Using Zero Product Rule


⇒ ( y + 2 ) = 0 and ( y - 5 ) = 0


⇒ y = - 2, 5


Now,


$<u>Case I : when y = - 2</u>$


x + y = 3


⇒ x - 2 = 3


⇒ x = 3 + 2


⇒ x = 5


$<u>Case II : when y = 5</u>$


x + y = 3


⇒ x + 5 = 3


⇒ x = 3 - 5


⇒ x = - 2



Hence, $<b>The zeroes are - 2 and 5.</b>$