solve this equation algebraicly and show work por favor!!!!!
x^2+y^2=25
3y-4x=0
thanks!!!!
2007-01-14 16:09:39 · 4 answers · asked by Fuzzyglasses 3 in Education & Reference ➔ Homework Help
Take the second equation and solve for x (or y)
4x = 3y
x = 3y /4 or x = 3/4 y
Then substitute that value into the first equation
x^2 + y^2 = 25
(3/4 y)^2 + y^2 = 25
9/16 y^2 + y^2 = 25
25/16 y^2 = 25
now divide both sides by 25/16
y^2 = 16
y = 4
now put the 4 in place of the y in the second equation
3*4 -4x = 0
12 - 4x = 0
x = 3
And it's always a good idea to check your answer by substituting your answers back into the original equations and see if they work.
x^2 + y^2 = 25
3^2 + 4^2 = 25
9 + 16 = 25
25 = 25
and
3y - 4x = 0
(3*4) - (4*3) = 0
12 - 12 = 0
0 = 0
Wow... it works!
2007-01-14 18:29:44 · answer #1 · answered by Irma R 2 · 1⤊ 0⤋
solve the 2nd equation for y, so 3y=4x, y=(4x)/3
now, plug that into the top equation and solve
x^2+(4x/3)^2=25
x^2+(16x^2/9)=25
multiply everything by 9 to get rid of the denominator
9x^2+16x^2=225
15x^2=225
x^2=15
x= + or - the (square root of)15
2007-01-15 00:22:37 · answer #2 · answered by kingsmansoysauce 2 · 0⤊ 1⤋
There are two correct answers... and Irma did a wonderful job of explaining.
2007-01-15 03:18:34 · answer #3 · answered by sagacity_ron 2 · 0⤊ 0⤋
y=4 x=3
there u go!
2007-01-15 00:22:55 · answer #4 · answered by tia 2 · 0⤊ 0⤋