X^2+Y^2=25
X+Y=7
Solve for X and Y algebraically.
X+Y=7
Solve for X and Y algebraically.
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The second equation gives
Y = 7-X
Substitute this into the first equation
X^2 + (7-X)^2 = 25
X^2 + 49 - 14X + X^2 = 25
2X^2 - 14X + 24 = 0
X^2 - 7X + 12 = 0
(X-3)(X-4) = 0
X = 3 or 4
Therefore
Y = 7-X
Y = 7-3 = 4
or
Y = 7-4 = 3
The solutions are (3,4) or (4,3)
Y = 7-X
Substitute this into the first equation
X^2 + (7-X)^2 = 25
X^2 + 49 - 14X + X^2 = 25
2X^2 - 14X + 24 = 0
X^2 - 7X + 12 = 0
(X-3)(X-4) = 0
X = 3 or 4
Therefore
Y = 7-X
Y = 7-3 = 4
or
Y = 7-4 = 3
The solutions are (3,4) or (4,3)
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X^2+Y^2=25
X+Y=7---------->transforms to---------->y = -x + 7
Solve for X and Y algebraically.
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This is a circle cut by, tangent to or not intersected by a line.
But, looking at the equations, the circle is centered at the origin, with a radius of 5.
Since the line has a -1 slope, and goes through y = 7(the y-intercept)
, the obvious conclusion is that the two equations do not intersect.
X+Y=7---------->transforms to---------->y = -x + 7
Solve for X and Y algebraically.
--------------------------------------…
This is a circle cut by, tangent to or not intersected by a line.
But, looking at the equations, the circle is centered at the origin, with a radius of 5.
Since the line has a -1 slope, and goes through y = 7(the y-intercept)
, the obvious conclusion is that the two equations do not intersect.
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x^2+y^x=25
x^2=25-y^2
x=√(25-y^2)
x+y=7
√(25-y^2)+y=7
√(25-y^2)=7-y
√(25-y^2)^2=(7-y)^2
25-y^2=(7-y)(7-y)
25-y^2=49-7y-7y+y^2
25-y^2=49-14y+y^2
25=49-14y+y^2+y^2
25=2y^2-14y+49
0=2y^2-14y+49-25
0=2y^2-14y+24
0=2(y^2-7y+12)
0/2=y^2-7y+12
0=y^2-7y+12
0=(y-4)(y-3)
Either (y-4)=0 or (y-3)=0
Therefore y=4 or y=3
You can plug in each value for y and solve for x. But you'll notice that you get x=3 when you plug in y=4, and you get x=4 when you plug in y=3.
Answer: x=3 and y=4 OR x=4 and y=3
x^2=25-y^2
x=√(25-y^2)
x+y=7
√(25-y^2)+y=7
√(25-y^2)=7-y
√(25-y^2)^2=(7-y)^2
25-y^2=(7-y)(7-y)
25-y^2=49-7y-7y+y^2
25-y^2=49-14y+y^2
25=49-14y+y^2+y^2
25=2y^2-14y+49
0=2y^2-14y+49-25
0=2y^2-14y+24
0=2(y^2-7y+12)
0/2=y^2-7y+12
0=y^2-7y+12
0=(y-4)(y-3)
Either (y-4)=0 or (y-3)=0
Therefore y=4 or y=3
You can plug in each value for y and solve for x. But you'll notice that you get x=3 when you plug in y=4, and you get x=4 when you plug in y=3.
Answer: x=3 and y=4 OR x=4 and y=3
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x + y = 7
y = 7 - x
x^2 + (7-x)^2 = 25
x^2 + x^2 - 14x + 49 = 25
2x^2 -14x + 24 = 0
2(x^2 -7x + 12) = 0
x=3 x=4
Okay, so at this point, the answers are going to be 3 and 4, just to let you know. I suppose you'd have to solve for Y too, but it would yield the same results. It doesn't matter whether X or Y is 3 or 4. You have no way of knowing. But yeah. 3 + 4 = 7. 3^2 + 4^2 = 25.
y = 7 - x
x^2 + (7-x)^2 = 25
x^2 + x^2 - 14x + 49 = 25
2x^2 -14x + 24 = 0
2(x^2 -7x + 12) = 0
x=3 x=4
Okay, so at this point, the answers are going to be 3 and 4, just to let you know. I suppose you'd have to solve for Y too, but it would yield the same results. It doesn't matter whether X or Y is 3 or 4. You have no way of knowing. But yeah. 3 + 4 = 7. 3^2 + 4^2 = 25.
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X² + Y²= 25
X + Y = 7 \\ solve for Y: Y = 7 - X and replace in other equation.
X² + (7 - X)² = 25 OR X² + X² - 14X + 49 = 25
2X² - 14X + 24 = 0 → X² - 7X + 12 = 0
This factors as (X - 4)(X - 3) = 0
X = 4 and Y = 3
OR
X = 3 and Y = 4
Check this result in the original equation(s), I did!
X + Y = 7 \\ solve for Y: Y = 7 - X and replace in other equation.
X² + (7 - X)² = 25 OR X² + X² - 14X + 49 = 25
2X² - 14X + 24 = 0 → X² - 7X + 12 = 0
This factors as (X - 4)(X - 3) = 0
X = 4 and Y = 3
OR
X = 3 and Y = 4
Check this result in the original equation(s), I did!
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X=3 or 4 9+16=25
Y=3 or 4 3+4=7
Apparently level 1 members aren't good enough to vote. Fook this.
Y=3 or 4 3+4=7
Apparently level 1 members aren't good enough to vote. Fook this.
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(X = 3 & Y = 4) , (X = 4 & Y = 3)