Solve (3x+y)/8 = (x-y)/5 = (x²-y²)/5. What is the values for x and y?

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Solve (3x+y)/8 = (x-y)/5 = (x²-y²)/5. What is the values for x and y?

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Answer 1 · 2015-07-04T03:10:07

The two solutions are: $(x, y) = (0, 0)$ $(x,y) = (0,0)$ and $(x, y) = (\frac{13}{6}, - \frac{7}{6})$ $(x,y) = (13/6, -7/6)$

Explanation:

$\frac{3 x + y}{8} = \frac{x - y}{5} = \frac{x^{2} - y^{2}}{5}$ $(3x+y)/8 = (x-y)/5 = (x^2-y^2)/5$

Start with $\frac{x - y}{5} = \frac{x^{2} - y^{2}}{5}$ $(x-y)/5 = (x^2-y^2)/5$ . Multiply by $5$ $5$ and factor the right side:

$(x - y) = (x - y) (x + y)$ $(x-y) = (x - y)(x+y)$ .

Collect on one side:
$(x - y) (x + y) - (x - y) = 0$ $(x - y)(x+y) - (x-y) = 0$ .

Factor $(x - y)$ $(x-y)$

$(x - y) (x + y - 1) = 0$ $(x - y)(x+y - 1) = 0$ .

So $x - y = 0$ $x-y=0$ or $x + y - 1 = 0$ $x+y-1 = 0$

This gives us: $y = x$ $y=x$ or $y = 1 - x$ $y = 1-x$

Now use the first two expressions together with these solutions for $y$ $y$ .

$\frac{3 x + y}{8} = \frac{x - y}{5}$ $(3x+y)/8 = (x-y)/5$
Leads to: $15 x + 5 y = 8 x - 8 y$ $15x+5y=8x-8y$ .

So $7 x + 13 y = 0$ $7x+13y =0$

Solution 1
Now, when $y = x$ $y=x$ , we get $20 x = 0$ $20x = 0$ , so $x = 0$ $x=0$ and thus $y = 0$ $y=0$

Solution 2
When $y = 1 - x$ $y=1-x$ , we get

$7 x + 13 (1 - x) = 0$ $7x+13(1-x)=0$

$7 x + 13 - 13 x = 0$ $7x + 13 -13x =0$

$- 6 x = - 13$ $-6x = -13$

$x = \frac{13}{6}$ $x=13/6$ and
$y = 1 - x = 1 - \frac{13}{6} = - \frac{7}{6}$ $y = 1-x = 1- 13/6 = -7/6$

Checking these solutions

$\frac{3 x + y}{8} = \frac{x - y}{5} = \frac{x^{2} - y^{2}}{5}$ $(3x+y)/8 = (x-y)/5 = (x^2-y^2)/5$

For $(0, 0)$ $(0,0)$ , we get

$\frac{0}{8} = \frac{0}{5} = \frac{0}{5}$ $0/8 = 0/5 =0/5$

For $(\frac{13}{6}, - \frac{7}{6})$ $(13/6, -7/6)$ , we get:

$\frac{3 (\frac{13}{6}) + (- \frac{7}{6})}{8} = \frac{39 - 7}{48} = \frac{32}{48} = \frac{2}{3}$ $(3(13/6)+(-7/6))/8 = (39-7)/48 = 32/48 = 2/3$
$\frac{(\frac{13}{6}) - (- \frac{7}{6})}{5} = \frac{20}{30} = \frac{2}{3}$ $((13/6)-(-7/6))/5 = 20/30 = 2/3$
$\frac{{(\frac{13}{6})}^{2} - {(- \frac{7}{6})}^{2}}{5} = \frac{169 - 49}{36 \cdot 5} = \frac{120}{36 \cdot 5} = \frac{20}{6 \cdot 5} = \frac{2}{3}$ $((13/6)^2-(-7/6)^2)/5 = (169 - 49)/(36*5) = 120/(36*5) = 20/(6*5) = 2/3$

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Solve (3x+y)/8 = (x-y)/5 = (x²-y²)/5. What is the values for x and y?

1 Answer

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