Exercise 5.4: System of Quadratic Equations with Two Variables
1. $(2x+3)(y-1) = 14, \quad (x-3)(y-2) = -1$
Solution: Given equations,
$(2x+3)(y-1) = 14 \quad \dots \dots (i)$
$(x-3)(y-2) = -1 \quad \dots \dots (ii)$
From equation $(i)$, we get,
$2xy – 2x + 3y – 3 = 14$
or, $2xy – 2x + 3y = 17 \quad \dots \dots (iii)$
From equation $(ii)$, we get,
$xy – 2x – 3y + 6 = -1$
or, $xy – 2x – 3y = -7 \quad \dots \dots (iv)$
Subtracting $(iv)$ from $(iii)$, we get:
$(2xy – 2x + 3y) – (xy – 2x – 3y) = 17 – (-7)$
or, $xy + 6y = 24$
or, $y(x + 6) = 24$
or, $y = \frac{24}{x + 6} \quad \dots \dots (v)$
Substituting the value of $y$ from $(v)$ into equation $(iv)$, we get,
$x\left(\frac{24}{x+6}\right) – 2x – 3\left(\frac{24}{x+6}\right) = -7$
or, $\frac{24x – 72}{x+6} – 2x = -7$
or, $\frac{24x – 72 – 2x(x+6)}{x+6} = -7$
or, $24x – 72 – 2x^2 – 12x = -7(x+6)$
or, $-2x^2 + 12x – 72 = -7x – 42$
or, $-2x^2 + 19x – 30 = 0$
or, $2x^2 – 19x + 30 = 0$
or, $2x^2 – 15x – 4x + 30 = 0$
or, $x(2x – 15) – 2(2x – 15) = 0$
or, $(2x – 15)(x – 2) = 0$
Either,
$2x – 15 = 0 \implies 2x = 15 \implies x = \frac{15}{2}$
Or,
$x – 2 = 0 \implies x = 2$
Now, if $x = \frac{15}{2}$, from $(v)$ we get:
$y = \frac{24}{\frac{15}{2} + 6} = \frac{24}{\frac{27}{2}} = \frac{48}{27} = \frac{16}{9}$
And if $x = 2$, from $(v)$ we get:
$y = \frac{24}{2+6} = \frac{24}{8} = 3$
Required solution:$(x, y) = (2, 3), \left(\frac{15}{2}, \frac{16}{9}\right)$
2. $(x-2)(y-1) = 3, \quad (x+2)(2y-5) = 15$
Solution: Given equations,
$(x-2)(y-1) = 3 \quad \dots \dots (i)$
$(x+2)(2y-5) = 15 \quad \dots \dots (ii)$
From equation $(i)$, we get:
$xy – x – 2y + 2 = 3$
or, $xy – x – 2y = 1 \quad \dots \dots (iii)$
From equation $(ii)$, we get,
$2xy – 5x + 4y – 10 = 15$
or, $2xy – 5x + 4y = 25 \quad \dots \dots (iv)$
Multiplying equation $(iii)$ by $2$ and subtracting from $(iv)$,
$(2xy – 5x + 4y) – 2(xy – x – 2y) = 25 – 2(1)$
or, $2xy – 5x + 4y – 2xy + 2x + 4y = 23$
or, $-3x + 8y = 23$
or, $8y – 23 = 3x$
or, $x = \frac{8y – 23}{3} \quad \dots \dots (v)$
Substituting the value of $x$ from equation $(v)$ into $(i)$, we get,
$\left(\frac{8y – 23}{3} – 2\right)(y – 1) = 3$
or, $\left(\frac{8y – 23 – 6}{3}\right)(y – 1) = 3$
or, $\frac{8y – 29}{3} \cdot (y – 1) = 3$
or, $(8y – 29)(y – 1) = 9$
or, $8y^2 – 8y – 29y + 29 = 9$
or, $8y^2 – 37y + 20 = 0$
or, $8y^2 – 32y – 5y + 20 = 0$
or, $8y(y – 4) – 5(y – 4) = 0$
or, $(y – 4)(8y – 5) = 0$
Either,
$y – 4 = 0 \implies y = 4$
Or,
$8y – 5 = 0 \implies y = \frac{5}{8}$
Now, if $y = 4$, from $(v)$ we get,
$x = \frac{8(4) – 23}{3} = \frac{32 – 23}{3} = \frac{9}{3} = 3$
And if $y = \frac{5}{8}$, from $(v)$ we get:
$x = \frac{8\left(\frac{5}{8}\right) – 23}{3} = \frac{5 – 23}{3} = \frac{-18}{3} = -6$
Required solution:$(x, y) = (3, 4), \left(-6, \frac{5}{8}\right)$
3. $x^2 = 7x + 6y, \quad y^2 = 7y + 6x$
Solution: Given equations,
$x^2 = 7x + 6y \quad \dots \dots (i)$
$y^2 = 7y + 6x \quad \dots \dots (ii)$
Subtracting $(ii)$ from $(i)$,
$x^2 – y^2 = (7x + 6y) – (7y + 6x)$
or, $x^2 – y^2 = x – y$
or, $(x – y)(x + y) – (x – y) = 0$
or, $(x – y)(x + y – 1) = 0$
Case One:
$x – y = 0 \implies x = y$
Substituting $x = y$ into $(i)$, we get,
$x^2 = 7x + 6x$
or, $x^2 = 13x$
or, $x^2 – 13x = 0$
or, $x(x – 13) = 0$
So, if $x = 0$, then $y = 0$;
and if $x = 13$, then $y = 13$.
Case Two:
$x + y – 1 = 0 \implies y = 1 – x$
Substituting $y = 1 – x$ into $(i)$, we get,
$x^2 = 7x + 6(1 – x)$
or, $x^2 = 7x + 6 – 6x$
or, $x^2 = x + 6$
or, $x^2 – x – 6 = 0$
or, $x^2 – 3x + 2x – 6 = 0$
or, $x(x – 3) + 2(x – 3) = 0$
or, $(x – 3)(x + 2) = 0$
Either $x = 3 \implies y = 1 – 3 = -2$
or $x = -2 \implies y = 1 – (-2) = 3$
Required solution:$(x, y) = (0, 0), (13, 13), (3, -2), (-2, 3)$
4. $x^2 = 3x + 2y, \quad y^2 = 3y + 2x$
Solution: Given equations,
$x^2 = 3x + 2y \quad \dots \dots (i)$
$y^2 = 3y + 2x \quad \dots \dots (ii)$
Subtracting $(ii)$ from $(i)$,
$x^2 – y^2 = (3x + 2y) – (3y + 2x)$
or, $x^2 – y^2 = x – y$
or, $(x – y)(x + y) – (x – y) = 0$
or, $(x – y)(x + y – 1) = 0$
Case One:
$x – y = 0 \implies x = y$
Substituting $x = y$ into $(i)$, we get,
$x^2 = 3x + 2x$
or, $x^2 = 5x$
or, $x^2 – 5x = 0$
or, $x(x – 5) = 0$
So, if $x = 0$, then $y = 0$;
and if $x = 5$, then $y = 5$.
Case Two:
$x + y – 1 = 0 \implies y = 1 – x$
Substituting $y = 1 – x$ into $(i)$, we get,
$x^2 = 3x + 2(1 – x)$
or, $x^2 = 3x + 2 – 2x$
or, $x^2 = x + 2$
or, $x^2 – x – 2 = 0$
or, $x^2 – 2x + x – 2 = 0$
or, $x(x – 2) + 1(x – 2) = 0$
or, $(x – 2)(x + 1) = 0$
Either $x = 2 \implies y = 1 – 2 = -1$
or $x = -1 \implies y = 1 – (-1) = 2$
Required solution:$(x, y) = (0, 0), (5, 5), (2, -1), (-1, 2)$
5. $x + \frac{4}{y} = 1, \quad y + \frac{4}{x} = 25$
Solution: Given equations,
$x + \frac{4}{y} = 1 \implies \frac{4}{y} = 1 – x \implies y = \frac{4}{1 – x} \quad \dots \dots (i)$
$y + \frac{4}{x} = 25 \quad \dots \dots (ii)$
Substituting the value of $y$ from equation $(i)$ into $(ii)$, we get,
$\frac{4}{1 – x} + \frac{4}{x} = 25$
or, $4\left(\frac{1}{1 – x} + \frac{1}{x}\right) = 25$
or, $4\left(\frac{x + 1 – x}{x(1 – x)}\right) = 25$
or, $4\left(\frac{1}{x – x^2}\right) = 25$
or, $\frac{4}{x – x^2} = 25$
or, $25(x – x^2) = 4$
or, $25x – 25x^2 = 4$
or, $25x^2 – 25x + 4 = 0$
or, $25x^2 – 20x – 5x + 4 = 0$
or, $5x(5x – 4) – 1(5x – 4) = 0$
or, $(5x – 4)(5x – 1) = 0$
Either $5x – 4 = 0 \implies x = \frac{4}{5}$
Or $5x – 1 = 0 \implies x = \frac{1}{5}$
Now, if $x = \frac{4}{5}$, from $(i)$ we get:
$y = \frac{4}{1 – \frac{4}{5}} = \frac{4}{\frac{1}{5}} = 20$
And if $x = \frac{1}{5}$, from $(i)$ we get:
$y = \frac{4}{1 – \frac{1}{5}} = \frac{4}{\frac{4}{5}} = 5$
Required solution:$(x, y) = \left(\frac{4}{5}, 20\right), \left(\frac{1}{5}, 5\right)$
6. $y + 3 = \frac{4}{x}, \quad x – 4 = \frac{5}{3y}$
Solution: Given equations,
$y + 3 = \frac{4}{x} \implies x = \frac{4}{y+3} \quad \dots \dots (i)$
$x – 4 = \frac{5}{3y} \quad \dots \dots (ii)$
Substituting the value of $x$ from $(i)$ into $(ii)$, we get,
$\frac{4}{y+3} – 4 = \frac{5}{3y}$
or, $\frac{4 – 4(y+3)}{y+3} = \frac{5}{3y}$
or, $\frac{4 – 4y – 12}{y+3} = \frac{5}{3y}$
or, $\frac{-4y – 8}{y+3} = \frac{5}{3y}$
or, $3y(-4y – 8) = 5(y + 3)$
or, $-12y^2 – 24y = 5y + 15$
or, $12y^2 + 29y + 15 = 0$
or, $12y^2 + 20y + 9y + 15 = 0$
or, $4y(3y + 5) + 3(3y + 5) = 0$
or, $(3y + 5)(4y + 3) = 0$
Either $3y + 5 = 0 \implies y = -\frac{5}{3}$
Or $4y + 3 = 0 \implies y = -\frac{3}{4}$
Now, if $y = -\frac{5}{3}$, from $(i)$ we get,
$x = \frac{4}{-\frac{5}{3} + 3} = \frac{4}{\frac{4}{3}} = 3$
And if $y = -\frac{3}{4}$, from $(i)$ we get:
$x = \frac{4}{-\frac{3}{4} + 3} = \frac{4}{\frac{9}{4}} = \frac{16}{9}$
Required solution:$(x, y) = \left(3, -\frac{5}{3}\right), \left(\frac{16}{9}, -\frac{3}{4}\right)$
7. $xy – x^2 = 1, \quad y^2 – xy = 2$
Solution: Given equations,
$xy – x^2 = 1 \implies x(y – x) = 1 \quad \dots \dots (i)$
$y^2 – xy = 2 \implies y(y – x) = 2 \quad \dots \dots (ii)$
Dividing equation $(ii)$ by equation $(i)$, we get,
$\frac{y(y – x)}{x(y – x)} = \frac{2}{1}$
or, $\frac{y}{x} = 2$
or, $y = 2x \quad \dots \dots (iii)$
Substituting $y = 2x$ into $(i)$, we get,
$x(2x) – x^2 = 1$
or, $2x^2 – x^2 = 1$
or, $x^2 = 1$
or, $x = \pm 1$
Now, if $x = 1$, then $y = 2(1) = 2$
And if $x = -1$, then $y = 2(-1) = -2$
Required solution:$(x, y) = (1, 2), (-1, -2)$
8. $x^2 – xy = 14, \quad y^2 + xy = 60$
Solution: Given equations,
$x^2 – xy = 14 \implies x(x – y) = 14 \quad \dots \dots (i)$
$y^2 + xy = 60 \implies y(x + y) = 60 \quad \dots \dots (ii)$
Dividing equation $(i)$ by equation $(ii)$, we get,
$\frac{x(x – y)}{y(x + y)} = \frac{14}{60} = \frac{7}{30}$
or, $30x(x – y) = 7y(x + y)$
or, $30x^2 – 30xy = 7xy + 7y^2$
or, $30x^2 – 37xy – 7y^2 = 0$
or, $30x^2 – 42xy + 5xy – 7y^2 = 0$
or, $6x(5x – 7y) + y(5x – 7y) = 0$
or, $(5x – 7y)(6x + y) = 0$
Case One:
$5x – 7y = 0 \implies x = \frac{7y}{5}$
Substituting the value of $x$ into $(ii)$, we get,
$y^2 + \left(\frac{7y}{5}\right)y = 60$
or, $y^2 + \frac{7y^2}{5} = 60$
or, $\frac{12y^2}{5} = 60$
or, $12y^2 = 300$
or, $y^2 = 25$
or, $y = \pm 5$
If $y = 5$, then $x = \frac{7(5)}{5} = 7$
If $y = -5$, then $x = \frac{7(-5)}{5} = -7$
Case Two:
$6x + y = 0 \implies y = -6x$
Substituting the value of $y$ into $(i)$, we get:
$x^2 – x(-6x) = 14$
or, $x^2 + 6x^2 = 14$
or, $7x^2 = 14$
or, $x^2 = 2$
or, $x = \pm \sqrt{2}$
If $x = \sqrt{2}$, then $y = -6\sqrt{2}$
If $x = -\sqrt{2}$, then $y = 6\sqrt{2}$
Required solution:$(x, y) = (7, 5), (-7, -5), (\sqrt{2}, -6\sqrt{2}), (-\sqrt{2}, 6\sqrt{2})$
9. $x^2 + y^2 = 25, \quad xy = 12$
Solution: Given equations,
$x^2 + y^2 = 25 \quad \dots \dots (i)$
$xy = 12 \quad \dots \dots (ii)$
We know,
$(x + y)^2 = x^2 + y^2 + 2xy = 25 + 2(12) = 25 + 24 = 49$
or, $x + y = \pm 7 \quad \dots \dots (iii)$
Also:
$(x – y)^2 = x^2 + y^2 – 2xy = 25 – 2(12) = 25 – 24 = 1$
or, $x – y = \pm 1 \quad \dots \dots (iv)$
Now, solving the combinations:
- Adding and subtracting $x + y = 7$ and $x – y = 1$ gives: $x = 4, y = 3$
- Adding and subtracting $x + y = 7$ and $x – y = -1$ gives: $x = 3, y = 4$
- Adding and subtracting $x + y = -7$ and $x – y = 1$ gives: $x = -3, y = -4$
- Adding and subtracting $x + y = -7$ and $x – y = -1$ gives: $x = -4, y = -3$
Required solution:$(x, y) = (4, 3), (3, 4), (-3, -4), (-4, -3)$
10. $\frac{x+y}{x-y} + \frac{x-y}{x+y} = \frac{10}{3}, \quad x^2 – y^2 = 3$
Solution: Given equations,
$\frac{x+y}{x-y} + \frac{x-y}{x+y} = \frac{10}{3} \quad \dots \dots (i)$
$x^2 – y^2 = 3 \quad \dots \dots (ii)$
Let, $\frac{x+y}{x-y} = a$.
Then equation $(i)$ becomes,
$a + \frac{1}{a} = \frac{10}{3}$
or, $\frac{a^2 + 1}{a} = \frac{10}{3}$
or, $3a^2 + 3 = 10a$
or, $3a^2 – 10a + 3 = 0$
or, $3a^2 – 9a – a + 3 = 0$
or, $3a(a – 3) – 1(a – 3) = 0$
or, $(a – 3)(3a – 1) = 0$
Either $a = 3$ or $a = \frac{1}{3}$.
Case One:$a = 3$
or, $\frac{x+y}{x-y} = 3$
or, $x + y = 3x – 3y$
or, $2x = 4y$
or, $x = 2y$
Substituting $x = 2y$ into $(ii)$, we get:
$(2y)^2 – y^2 = 3$
or, $4y^2 – y^2 = 3$
or, $3y^2 = 3$
or, $y^2 = 1 \implies y = \pm 1$
If $y = 1 \implies x = 2(1) = 2$
If $y = -1 \implies x = 2(-1) = -2$
Case Two: $a = \frac{1}{3}$
or, $\frac{x+y}{x-y} = \frac{1}{3}$
or, $3x + 3y = x – y$
or, $2x = -4y$
or, $x = -2y$
Substituting $x = -2y$ into $(ii)$, we get,
$(-2y)^2 – y^2 = 3$
or, $4y^2 – y^2 = 3$
or, $3y^2 = 3$
or, $y^2 = 1 \implies y = \pm 1$
If $y = 1 \implies x = -2(1) = -2$
If $y = -1 \implies x = -2(-1) = 2$
Required solution:$(x, y) = (2, 1), (-2, -1), (-2, 1), (2, -1)$
11. $x^2 + xy + y^2 = 3, \quad x^2 – xy + y^2 = 7$
Solution: Given equations,
$x^2 + xy + y^2 = 3 \quad \dots \dots (i)$
$x^2 – xy + y^2 = 7 \quad \dots \dots (ii)$
Adding $(i)$ and $(ii)$,
$2(x^2 + y^2) = 10 \implies x^2 + y^2 = 5 \quad \dots \dots (iii)$
Subtracting $(ii)$ from $(i)$,
$2xy = -4 \implies xy = -2 \quad \dots \dots (iv)$
Now,
$(x + y)^2 = x^2 + y^2 + 2xy = 5 + 2(-2) = 5 – 4 = 1 \implies x + y = \pm 1 \quad \dots \dots (v)$
$(x – y)^2 = x^2 + y^2 – 2xy = 5 – 2(-2) = 5 + 4 = 9 \implies x – y = \pm 3 \quad \dots \dots (vi)$
Solving for different values:
- $x + y = 1$ and $x – y = 3$ gives: $x = 2, y = -1$
- $x + y = 1$ and $x – y = -3$ gives: $x = -1, y = 2$
- $x + y = -1$ and $x – y = 3$ gives: $x = 1, y = -2$
- $x + y = -1$ and $x – y = -3$ gives: $x = -2, y = 1$
Required solution:$(x, y) = (2, -1), (-1, 2), (1, -2), (-2, 1)$
12. $2x^2 + 3xy + y^2 = 20, \quad 5x^2 + 4y^2 = 41$
Solution: Given equations,
$2x^2 + 3xy + y^2 = 20 \quad \dots \dots (i)$
$5x^2 + 4y^2 = 41 \quad \dots \dots (ii)$
Multiplying $(i)$ by $41$ and $(ii)$ by $20$, then subtracting,
$41(2x^2 + 3xy + y^2) – 20(5x^2 + 4y^2) = 0$
or, $82x^2 + 123xy + 41y^2 – 100x^2 – 80y^2 = 0$
or, $-18x^2 + 123xy – 39y^2 = 0$
or, $-3(6x^2 – 41xy + 13y^2) = 0$
or, $6x^2 – 41xy + 13y^2 = 0$
or, $6x^2 – 39xy – 2xy + 13y^2 = 0$
or, $3x(2x – 13y) – y(2x – 13y) = 0$
or, $(2x – 13y)(3x – y) = 0$
Case One:
$3x – y = 0 \implies y = 3x$
Substituting $y = 3x$ into $(ii)$, we get,
$5x^2 + 4(3x)^2 = 41$
or, $5x^2 + 4(9x^2) = 41$
or, $5x^2 + 36x^2 = 41$
or, $41x^2 = 41 \implies x^2 = 1 \implies x = \pm 1$
If $x = 1 \implies y = 3(1) = 3$
If $x = -1 \implies y = 3(-1) = -3$
Case Two:
$2x – 13y = 0 \implies x = \frac{13y}{2}$
Substituting $x = \frac{13y}{2}$ into $(ii)$, we get,
$5\left(\frac{13y}{2}\right)^2 + 4y^2 = 41$
or, $5\left(\frac{169y^2}{4}\right) + 4y^2 = 41$
or, $\frac{845y^2}{4} + 4y^2 = 41$
or, $\frac{845y^2 + 16y^2}{4} = 41$
or, $\frac{861y^2}{4} = 41$
or, $861y^2 = 164 \implies y^2 = \frac{164}{861} = \frac{4}{21} \implies y = \pm \frac{2}{\sqrt{21}}$
If $y = \frac{2}{\sqrt{21}} \implies x = \frac{13}{2} \cdot \frac{2}{\sqrt{21}} = \frac{13}{\sqrt{21}}$
If $y = -\frac{2}{\sqrt{21}} \implies x = -\frac{13}{\sqrt{21}}$
Required solution:$(x, y) = (1, 3), (-1, -3), \left(\frac{13}{\sqrt{21}}, \frac{2}{\sqrt{21}}\right), \left(-\frac{13}{\sqrt{21}}, -\frac{2}{\sqrt{21}}\right)$