y= 2x - 5
x(squared) + y(squared) = 25
the answers are:
x = 0
x = 4
y= -5
y= 3
can someone explain how you do this? thanks
how do i solve these simultaneous equations?
2008-05-17 2:25 pm
回答 (9)
2008-05-17 2:31 pm
✔ 最佳答案
y = 2x - 5x² - y² = 25
Substitute first equation into second equation:
x² + (2x - 5)² = 25
Expand second term on left:
x² + (4x² - 10x - 10x + 25) = 25
Combine like terms:
5x² - 20x + 25 = 25
Subtract 25 from both sides:
5x² - 20x = 0
Factor out an x:
x(5x - 20) = 0
By the zero product principle:
x = 0, 5x - 20 = 0
x = 0, 5x = 20
x = 0, x = 4
When x = 0:
y = 2*0 - 5 = -5
When x = 4:
y = 2*4 - 5 = 3
The solutions are:
(0, - 5); (4, 3)
2008-05-17 9:35 pm
ok so lets call the first equation 1 and the second one 2
ok so sub the first equation into second one, since y = 2x-5 you replace the y in the other equation with this.
So x^2+(2x-5)^2=25
foil it out x^2+4x^20x+25=25
ok next collect terms and make one side = 0
so 5x^2-20x = 0 now you factorise and use the null factor
so 5x(x-4)= 0
so x = 0 (first x)
what inside the brackets u need to make 0 = 4
so x = 0 or x = 4 next
sub back in
so you go y = 2x 0 -5
y = -5
y = 2x8 -5
so y = 3
soooo x = 0 x = 4 and y = -5 y = 3
EDIT: god dam, people beat me to it while i was writing this out...sigh
ok so sub the first equation into second one, since y = 2x-5 you replace the y in the other equation with this.
So x^2+(2x-5)^2=25
foil it out x^2+4x^20x+25=25
ok next collect terms and make one side = 0
so 5x^2-20x = 0 now you factorise and use the null factor
so 5x(x-4)= 0
so x = 0 (first x)
what inside the brackets u need to make 0 = 4
so x = 0 or x = 4 next
sub back in
so you go y = 2x 0 -5
y = -5
y = 2x8 -5
so y = 3
soooo x = 0 x = 4 and y = -5 y = 3
EDIT: god dam, people beat me to it while i was writing this out...sigh
2008-05-18 12:30 am
x² + (2x - 5)² = 25
x² + 4x² - 20x + 25 = 25
5x² - 20x = 0
5x (x - 4) = 0
x = 0 , x = 4
y = - 5 , y = 3
x² + 4x² - 20x + 25 = 25
5x² - 20x = 0
5x (x - 4) = 0
x = 0 , x = 4
y = - 5 , y = 3
2008-05-17 11:52 pm
(1) y=2x-5
(2) x²+y²=25
substitute equation (1) in equation (2) :
x²+(2x-5)²=25
then multiply out the bracket (2x-5)² which gives you 4x²-20x+25. Then:
x²+4x²-20x+25=25
5x²-20x+25=25
bring the 25 to the other side, so it equals 0 :
5x²-20x=0
now take 5x out as a common factor:
5x(x-4)=0
for that equation to equal zero, either the quantity inside the bracket or the 5x must be zero.
for the quantity inside the bracket to be zero, the "x" value must be 4 (because 4-4=0).
For the "5x" to be zero, the "x" value must be zero (anything multiplied by zero is zero).
Now you know the two possible values for "x" :
x=4
x=0
Now substitute these values in (1)
y=2x-5
for x=0
y=2*0-5
y=0-5
y=-5
for x=4
y=2x-5
y=2*4-5
y=8-5
y=3
So when x=4, y= 3
and when x=0, y=-5
(2) x²+y²=25
substitute equation (1) in equation (2) :
x²+(2x-5)²=25
then multiply out the bracket (2x-5)² which gives you 4x²-20x+25. Then:
x²+4x²-20x+25=25
5x²-20x+25=25
bring the 25 to the other side, so it equals 0 :
5x²-20x=0
now take 5x out as a common factor:
5x(x-4)=0
for that equation to equal zero, either the quantity inside the bracket or the 5x must be zero.
for the quantity inside the bracket to be zero, the "x" value must be 4 (because 4-4=0).
For the "5x" to be zero, the "x" value must be zero (anything multiplied by zero is zero).
Now you know the two possible values for "x" :
x=4
x=0
Now substitute these values in (1)
y=2x-5
for x=0
y=2*0-5
y=0-5
y=-5
for x=4
y=2x-5
y=2*4-5
y=8-5
y=3
So when x=4, y= 3
and when x=0, y=-5
2008-05-17 9:41 pm
y = 2x - 5
x^2 + y^2 = 25
y = 2x - 5
x^2 + y^2 = 25
x^2 + (2x - 5)^2 = 25
x^2 + (2x - 5)(2x - 5) = 25
x^2 + 4x^2 - 20x + 25 = 25
5x^2 - 20x + 25 - 25 = 0
5x^2 - 20x = 0
x^2 - 4x = 0
x(x - 4) = 0
x = 0
x - 4 = 0
x = 4
y = 2x - 5
y = 2(0) - 5
y = 0 - 5
y = -5
y = 2x - 5
y = 2(4) - 5
y = 8 - 5
y = 3
â´ (x = 0 , y = -5) , (x = 4 , y = 3)
x^2 + y^2 = 25
y = 2x - 5
x^2 + y^2 = 25
x^2 + (2x - 5)^2 = 25
x^2 + (2x - 5)(2x - 5) = 25
x^2 + 4x^2 - 20x + 25 = 25
5x^2 - 20x + 25 - 25 = 0
5x^2 - 20x = 0
x^2 - 4x = 0
x(x - 4) = 0
x = 0
x - 4 = 0
x = 4
y = 2x - 5
y = 2(0) - 5
y = 0 - 5
y = -5
y = 2x - 5
y = 2(4) - 5
y = 8 - 5
y = 3
â´ (x = 0 , y = -5) , (x = 4 , y = 3)
2008-05-17 9:37 pm
y = 2x - 5__(1)
x² + y² = 25__(2)
Substitute equation (1) into (2);
x² + (2x - 5)² = 25
x² + 4x² - 20x + 25 = 25
5x² - 20x = 0
5(x² - 4x) = 0
x² - 4x = 0
x(x - 4) = 0
x = 0 or x = 4
therefore;
y = 2(0) - 5
= -5
...OR...
y = 2(4) - 5
= 8 - 5
= 3
x² + y² = 25__(2)
Substitute equation (1) into (2);
x² + (2x - 5)² = 25
x² + 4x² - 20x + 25 = 25
5x² - 20x = 0
5(x² - 4x) = 0
x² - 4x = 0
x(x - 4) = 0
x = 0 or x = 4
therefore;
y = 2(0) - 5
= -5
...OR...
y = 2(4) - 5
= 8 - 5
= 3
2008-05-17 9:33 pm
1st, u have to subs. the value of y into the second eq.
x^2 + (2x-5)^2 =25
x^2 + 4x^2 - 20x + 25 =25
x^2 + 4x^2 - 20x = 0
5x^2 - 20x=0
x(5x - 20) =0
x =0, x=20/5 = 4
Then u just need to subs. the value of x into the 1st eq.
if u still don't understand it u can me msg me at ym ( guitargirlnameblackm)
x^2 + (2x-5)^2 =25
x^2 + 4x^2 - 20x + 25 =25
x^2 + 4x^2 - 20x = 0
5x^2 - 20x=0
x(5x - 20) =0
x =0, x=20/5 = 4
Then u just need to subs. the value of x into the 1st eq.
if u still don't understand it u can me msg me at ym ( guitargirlnameblackm)
2008-05-17 9:33 pm
y = 2x - 5
plug this into the second equation
x^2 + y^2 = 25
x^2 + (2x - 5)^2 = 25
Foil and add like terms
x^2 + 4x^2 - 20x + 25 = 25
5x^2 - 20x = 0
Factor:
5x(x-4) = 0
5x = 0 or x - 4 = 0
x = 0 or x = 4
Now for y, plug in your "x" values into the first equation
y = 2x - 5
y = 2(0) - 5
y = -5
y = 2x - 5
y = 2(4) - 5
y = 8 - 5
y = 3
plug this into the second equation
x^2 + y^2 = 25
x^2 + (2x - 5)^2 = 25
Foil and add like terms
x^2 + 4x^2 - 20x + 25 = 25
5x^2 - 20x = 0
Factor:
5x(x-4) = 0
5x = 0 or x - 4 = 0
x = 0 or x = 4
Now for y, plug in your "x" values into the first equation
y = 2x - 5
y = 2(0) - 5
y = -5
y = 2x - 5
y = 2(4) - 5
y = 8 - 5
y = 3
2008-05-17 9:29 pm
substitute the 1st one into the 2nd one
solve for x (you will get 2 answers)
plug those answers into the 1st equation to find y
what you are trying to do is find out where the line crosses the circle
solve for x (you will get 2 answers)
plug those answers into the 1st equation to find y
what you are trying to do is find out where the line crosses the circle
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