2x + y = 7
x2 -xy = 6
2006-12-24 18:07:25 · 12 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
2x + y = 7- - - - - -Equation 1
x² + xy = 6-- - - - -Equation 2
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Substitute Method equation 1
2x + y = 7
2x + y - 2x = 7 - 2x
y = 7 - 2x
The answer is y = 7 - 2x
insert the y value into equation 2
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x² - xy = 6
x² - x(7 - 2x) = 6
x² - 7x + 2x² = 6
3x² - 7x = 6
3x² - 7x - 6 = 6 - 6
3x² - 7x - 6 = 0
3x² - 9x + 2x - 6 = 0
3x(x - 3) + 2(x - 3) = 0
(3x + 2)(x - 3) = 0
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Roots
3x + 2 = 0
3x + 2 - 2 = 0 - 2
3x = - 2
3x/3 = - 2 / 3
x = - 2 / 3. . .<=. .not used
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x - 3 = 0
x - 3 + 3 = 0 + 3
x = 3
The answer is x = 3
Insert the x value into equation 1
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2x + y = 7
2(3) + y = 7
6 + y = 7
6 + y - 6 = 7 - 6
y = 1
The answer is y = 1
Insert the y value into equation 1
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Check for equation 1
2x + y = 7
2(3) + 1 = 7
6 + 1 = 7
7 = 7
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Check for equation 2
x² - xy = 6
(3)² - 3(1) = 6
9 - 3 = 6
6 = 6
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The solution set is { 3, 1 }
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2006-12-25 05:54:25 · answer #1 · answered by SAMUEL D 7 · 1⤊ 0⤋
2x + y = 7
x^2 - xy = 6
To solve this system of equations, the best way to approach this is substitution. For the first equation, 2x + y = 7, note that we can solve for y and get y = 7 - 2x.
Now, we take this value of y = 7 - 2x and plug it into the second equation.
x^2 - xy = 6
x^2 - x(7 - 2x) = 6. Expanding, we get
x^2 - 7x + 2x^2 = 6. Grouping together like terms,
3x^2 - 7x = 6. Moving everything to the left hand side,
3x^2 - 7x - 6 = 0.
Through trial and error, we see that this factors into
(3x + 2)(x - 3) = 0, which means
3x + 2 = 0
x - 3 = 0
Solving for x in each equation, we get the following values:
x = {-2/3, 3}
Note that we have TWO values for x. This means we plug in each value individually into the first equation to get values for y.
Since we already solved y in terms of x, let's use this as our guidelines.
y = 7 - 2x
If x = -2/3 :
y = 7 - 2(-2/3) = 7 + 4/3 = 21/3 + 4/3 = 25/3
If x = 3 :
y = 7 - 2(3) = 7 - 6 = 1
Therefore, our solutions are:
{x = -2/3, y = 25/3}
{x = 3, y = 1}
2006-12-25 02:16:21 · answer #2 · answered by Puggy 7 · 4⤊ 0⤋
I knew there was a mistake in your writing of the question.
From the first equation y = 7-2x.
Substitute in the second equation:
x^2-x(7-2x) =6
x^2+2x^2-7x = 6
3x^2 -7x -6 =0
(3x+2)(x-3) = 0
x =3 then y = 7-2(3) = 1 or
x = - 2/3 then y = 7-2(-2/3) = 25/3
Thus, the solutions are (3,1)&(-2/3,25/3)
2006-12-25 02:29:19 · answer #3 · answered by mulla sadra 3 · 2⤊ 0⤋
On the first equation, you need to subtract 2x from both sides:
y=-2x+7
Now, you can substitute it into the second equation:
x^2-x(-2x+7)=6
x^2+2x^2-7x=6
3x^2-7x-6=0
(3x+2)(x-3)=0
x=-2/3 and 3
2(-2/3)+y=7
-4/3+y=7
y=21/3+4/3
y=25/3
2(3)+y=7
6+y=7
y=1
y=25/3 and 1
(-2/3,25/3) and (3,1)
Check:
2(3)+1=7
6+1=7
7=7
(3)^2-(3*1)=6
9-3=6
6=6
2(-2/3)+25/3=7
-4/3+25/3=7
21/3=7
7=7
(-2/3)^2-(-2/3*25/3)=6
4/9+50/9=6
54/9=6
6=6
2006-12-25 02:32:03 · answer #4 · answered by Anonymous · 2⤊ 0⤋
2x + y = 7
x^2 -xy = 6
X^2 – x(7-2x)= 6
X^2 – 7x +2x^2 = 6
3x^2 -7x -6 = 0
3x^2 –(9 - 2)x - 6 = 0
3x^2 –9x + 2x - 6 = 0
3x(x – 3) + 2 (x – 3) = 0
(x - 3)(3x + 2) = 0
x – 3 = 0 or 3x + 2 = 0
x = 3 or x = -2/3
if x = -2/3, y = 8+1/3 = 25/3
if x = 3, y = 1
2006-12-25 02:24:03 · answer #5 · answered by Kinu Sharma 2 · 3⤊ 1⤋
Use combination,
2x + y = 7......(1)
x^2 -xy = 6......(2)
(1)x +(2):
3x^2 - 7x - 6 = 0......(3)
Factor (3):
(3x+2)(x-3) = 0......(4)
Therefore, the two solutions are
(-2/3, 25/3) and (3, 1)
2006-12-25 02:23:25 · answer #6 · answered by sahsjing 7 · 3⤊ 0⤋
2x + y = 7
y = 7-2x
Substitute in x^2 -xy = 6
x^2 - x(7-2x) = 6
x^2 - 7x + 2x^2 = 6
3x^2 - 7x - 6 = 0
3x^2 - 9x + 2x - 6 = 0
3x(x-3) + 2(x-3) = 0
(x-3)(3x+2) = 0
x=3 or x=-2/3
y = 7-2*3 or y = 7+2*2/3
y=1 or y = 25/3
Solution Sets are
{x=3,y=1}
and
{x=(-2/3),y=(25/3)}
2006-12-25 02:10:25 · answer #7 · answered by Som™ 6 · 3⤊ 0⤋
assuming that the 2nd equation is "x times 2" and not some attempt to type out a square.
solve for y in 1st Equation y=-2x+7
when put in the 2nd equation 2x-x(-2x+7)=6 yields 2x+2x^(2)-7x=6
when you bring over the 6 and combine like terms you get 2x^(2)-5x-6
you use quadratic equation to solve for the x's
(5+â((-5)^(2)-4(2)(-6)))/(2)(2) = 3.386 = x
(5-â((-5)^(2)-4(2)(-6)))/(2)(2) = -0.886 = x
put those two answers into any of the original equations
2x+y=7 modified to -2x+7=y
-2(3.386)+7= y = 0.228
-2(-0.886)+7= y = 8.772
So your two answers for this system of equations are(x, y)
(3.386, 0.228) and (-0.886, 8.772)
2006-12-25 03:35:13 · answer #8 · answered by Charlie-chan 1 · 1⤊ 0⤋
x2 -xy = 6... do you mean x^2 - xy = 6?
If so, y = 7-2x
x^2 - x(7-2x) = 6
3x^2 - 7x - 6 = 0
(3x + 2)(x - 3) = 0
x = -2/3 or 3
if x = -2/3, y = 8+1/3
if x = 3, y = 1
2006-12-25 02:17:12 · answer #9 · answered by buaya123 3 · 3⤊ 0⤋
The answer of this Question which iam getting is
16x + (y X y) - 42 = 0
2006-12-25 04:05:28 · answer #10 · answered by Praraj Ranka 1 · 1⤊ 0⤋