Solutions of Exercise 2.7 Class 10 Complete Guide
In Exercise 2.7 Class 10 Mathematics, we explore techniques for solving simultaneous equations in two variables when the system involves a mix of linear and quadratic equations. This skill is crucial, as it not only deepens our understanding of algebra but also applies to many practical situations in mathematics and physics, where systems of equations help describe relationships between variables in real-world problems.
Types of Systems in Exercise 2.7 Class 10
- Linear-Quadratic System
In this type, one equation is linear, while the other is quadratic. We can solve this system by substituting the linear equation into the quadratic equation, reducing the problem to a single variable. Once we solve for one variable, we can substitute back to find the other variable. - Quadratic-Quadratic System
Here, both equations are quadratic. To solve such a system, we typically use substitution or elimination to combine the equations and reduce them to a simpler form. Solving these systems often requires techniques for working with quadratic equations, such as factoring or applying the quadratic formula.
Approaches to Solving Systems of Equations
For both types of equation in exercise 2.7 class 10, the general approach involves:
- Substitution: Replace one variable from one equation into the other, simplifying the system to a single equation.
- Elimination: If substitution is challenging, we can align the equations to eliminate one variable, reducing the system to an easier-to-solve equation.
- Checking for Real Solutions: Since quadratic equations can sometimes yield complex numbers, we also assess the nature of the solutions to determine whether real solutions exist.
In the following solutions of exercise 2.7 class 10, you’ll see each step broken down, making it clear how to approach both linear-quadratic and quadratic-quadratic systems. Practice is key, so work through these solutions carefully to gain confidence with each type of system.
Question 1 of Exercise 2.7 Class 10. \(\boldsymbol{( x + y = 5 ) and ( x^2 – 2y = 14 )}\)
\(\text{From } x + y = 5, \text{ we get } x = 5 – y \\
\text{Substitute } x = 5 – y \text{ into } x^2 – 2y = 14: \\
(5 – y)^2 – 2y = 14 \\
25 – 10y + y^2 – 2y = 14 \\
y^2 – 12y + 11 = 0 \\
\text{Factoring: } (y – 11)(y – 1) = 0 \\
y = 11 \text{ or } y = 1 \\
\text{For } y = 11, \quad x + 11 = 5 \Rightarrow x = -6 \\
\text{For } y = 1, \quad x + 1 = 5 \Rightarrow x = 4 \\
\text{Solution Set: } \{(-6, 11), (4, 1)\}
\)
Question 2 of Exercise 2.7 Class 10. \(\boldsymbol{( 3x – 2y = 1 ) and ( x^2 + xy – y^2 = 1 )}\)
\(\text{From } 3x – 2y = 1, \text{ we get } x = \frac{1 + 2y}{3} \\
\text{Substitute this into } x^2 + xy – y^2 = 1: \\
\left( \frac{1 + 2y}{3} \right)^2 + \left( \frac{1 + 2y}{3} \right)y – y^2 = 1 \\
\frac{1 + 4y + 4y^2}{9} + \frac{y + 2y^2}{3} – y^2 = 1 \\
1 + 7y = 9 \Rightarrow y = \frac{8}{7} \\
\text{Substitute } y = \frac{8}{7} \text{ into } x = \frac{1 + 2y}{3}: \\
x = \frac{1 + \frac{16}{7}}{3} = \frac{23}{21}
\)
Question 3 of Exercise 2.7 Class 10. \(\boldsymbol{( x – y = 7 ) and ( \frac{2}{x} + \frac{5}{y} = 2 )}\)
\(\text{Multiplying both sides of } \frac{2}{x} + \frac{5}{y} = 2 \text{ by } xy, \text{ we get } 2y – 5x = 2xy \\
\text{From } x – y = 7, \text{ substitute } x = 7 + y \text{ into the equation: } \\
2y – 35 – 5y = 14y + 2y^2 \\
2y^2 + 17y + 35 = 0 \\
(y + 5)(2y + 7) = 0 \\
y = -5 \text{ or } y = -\frac{7}{2} \\
\text{For } y = -5, \quad x = 2 \\
\text{Now putting these values of } y \text{ in equation (i)} \\
\begin{array}{l|l}\\
\text{When } y = -5 & \text{When } y = \frac{-7}{2} \\
x = 7 + y & x = 7 + y \\
x = 7 + (-5) & x = 7 + \left( \frac{-7}{2} \right) \\
x = 7 – 5 & x = 7 – \frac{7}{2} \\
x = 2 & x = \frac{14 – 7}{2} \\
& x = \frac{7}{2} \\
\end{array}\\
\text{Solution Set is } \left\{ (2, -5), \left( \frac{7}{2}, \frac{-7}{2} \right) \right\}\\
\)
Question 4 of Exercise 2.7 Class 10. \(\boldsymbol{ x + y = a – b } \)
\(\frac{a}{x} – \frac{b}{y} = 2 \\
\textbf{Solution:} \\
x + y = a – b \quad \text{(i)} \\
\frac{a}{x} – \frac{b}{y} = 2 \Rightarrow \frac{ay – bx}{xy} = 2 \\
ay – bx = 2xy \quad \text{(ii)} \\
\text{From equation (i)} \\
x = a – b – y \quad \text{(iii)} \\
\text{Put it in equation (ii)}: \\
ay – bx = 2xy \\
ay – b(a – b – y) = 2(a – b – y)y \\
ay – ba + b^2 + by = 2ay – 2by – 2y^2 \\
2y^2 – 2ay + ay + 2by + by + b^2 – ab = 0 \\
2y^2 – ay + 3by + b^2 – ab = 0 \\
2y^2 – y(a – 3b) + (b^2 – ab) = 0 \\
\text{By using quadratic formula} \\
a = 2, \quad b = -(a – 3b), \quad c = (b^2 – ab) \\
y = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a} \\
y = \frac{-(a – 3b) \pm \sqrt{(a – 3b)^2 – 4(2)(b^2 – ab)}}{2(2)} \\
y = \frac{(a – 3b) \pm \sqrt{a^2 + 9b^2 – 6ab – 8b^2 + 8ab}}{4} \\
y = \frac{(a – 3b) \pm \sqrt{a^2 + b^2 + 2ab}}{4} \\
y = \frac{(a – 3b) \pm (a + b)}{4} \\
\text{Thus, either} \\
y = \frac{-4b}{4} \quad \text{or} \quad y = \frac{2a – 2b}{4} \\
y = -b \quad \text{or} \quad y = \frac{a – b}{2} \\
\text{Putting these values of } y \text{ in equation (iii)}: \\
\text{When } y = -b: \\
x = a – b – y \\
x = a – b – (-b) \\
x = a – b + b \\
x = a \\
\text{When } y = \frac{a – b}{2}: \\
x = a – b – y \\
x = a – b – \frac{a – b}{2} \\
x = \frac{2a – 2b – a + b}{2} \\
x = \frac{a – b}{2} \\
\text{Solution Set is } \left\{ (a, -b), \left( \frac{a – b}{2}, \frac{a – b}{2} \right) \right\}
\)
Question 5 of Exercise 2.7 Class 10. \(x^2 + (y – 1)^2= 10 \) \(\text{ and }\) \(x^2 + y^2 + 4x= 1 \)
\begin{aligned}
x^2 + (y – 1)^2 &= 10 \quad \text{(i)} \\
x^2 + y^2 + 4x &= 1 \quad \text{(ii)}
\end{aligned}
\(\textbf{Solution:} \)
\begin{aligned}
&x^2 + (y – 1)^2 = 10 \quad \text{(i)} \\
&x^2 + y^2 + 4x = 1 \quad \text{(ii)} \\
&\text{Subtracting equation (ii) from (i):} \\
&x^2 + y^2 + 1 – 2y = 10 \\
&x^2 + y^2 + 4x = 1 \\
&\pm x^2 + y^2 + 4x = \pm 1 \\
&-4x – 2y = 9 \\
&-4x – 2y = 9 – 1 \\
&-4x – 2y = 8 \\
&-2(2x + y) = 8 \\
&\Rightarrow 2x + y = \frac{8}{-2} \\
&2x + y = -4 \quad \text{(iii)} \\
&y = -4 – 2x \quad \text{(iii)} \\
&\text{Put in equation (ii):} \\
&x^2 + (-4 – 2x)^2 + 4x = 1 \\
&x^2 + \left[(-4 – 2x)^2\right] + 4x = 1 \\
&x^2 + \left[16 + 4x^2 + 16x\right] + 4x = 1 \\
&5x^2 + 20x + 15 = 0 \\
&x(x + 3)(x + 1) = 0 \\
&\text{Either} \, x + 3 = 0 \, \text{or} \, x + 1 = 0 \\
&x = -3 \, \text{or} \, x = -1
\end{aligned}
\(\text{Putting these values of \( x \) in equation (iii):} \)
\begin{aligned}
&\text{When } x = -3 \Rightarrow y = -4 – 2x \\
&y = -4 – 2(-3) \Rightarrow y = 2 \\
&\text{When } x = -1 \Rightarrow y = -4 – 2(-1) \\
&y = 0 \\
&\text{So, the solution set is: } \{(-3, 2), (-1, -2)\}
\end{aligned}
Question 6 of Exercise 2.7 Class 10. \( (x + 1)^2 + (y + 1)^2 = 5 \) \(\text{ and }\) \( (x + 2)^2 + y^2 = 5 \)
\begin{aligned}
(x + 1)^2 + (y + 1)^2 &= 5 \quad \text{(i)} \\
(x + 2)^2 + y^2 &= 5 \quad \text{(ii)}
\end{aligned}
\(\textbf{Solution:} \)
\begin{aligned}
& (x + 1)^2 + (y + 1)^2 = 5 \quad \text{(i)} \\
& (x + 2)^2 + y^2 = 5 \quad \text{(ii)} \\
& \text{From equation (i):} \\
& x^2 + 1 + 2x + y^2 + 1 + 2y = 5 \\
& x^2 + y^2 + 2x + 2y = 3 \quad \text{(iii)} \\
& \text{From equation (ii):} \\
& x^2 + y^2 + 4x = 1 \quad \text{(iv)} \\
& \text{Subtracting equation (iv) from (iii):} \\
& x^2 + y^2 + 2x + 2y = 3 \\
& x^2 + y^2 + 4x = 1 \\
& \pm x^2 + y^2 + 4x = \pm 1 \\
& -2(x – y) = 2 \\
& x – y = -1 \Rightarrow x = y – 1 \quad \text{(v)} \\
& \text{Putting } x = y – 1 \text{ in equation (iv):} \\
& (y – 1 + 2)^2 + y^2 = 5 \\
& (y + 1)^2 + y^2 = 5 \\
& y^2 + 2y + 1 + y^2 = 5 \\
& 2y^2 + 2y – 4 = 0 \\
& y(y + 2) – (y + 2) = 0 \Rightarrow (y + 2)(y – 1) = 0 \\
& \text{Either} \, y + 2 = 0 \, \text{or} \, y – 1 = 0 \\
& y = -2 \, \text{or} \, y = 1
\end{aligned}
\(\text{Putting these values of \( y \) in equation (v):} \)
\begin{aligned}
&\text{When } y = -2 \Rightarrow x = y – 1 \\
&x = -2 – 1 = -3 \\
&\text{When } y = 1 \Rightarrow x = y – 1 \\
&x = 1 – 1 = 0 \\
&\text{So, the solution set is: } \{(-3, -2), (0, 1)\}
\end{aligned}
Question 7 of Exercise 2.7 Class 10. \(( x^2 + 2y^2= 22) \text{ and } (5x^2 + y^2 = 29) \)
\begin{aligned}
x^2 + 2y^2 &= 22 \quad \text{(i)} \\
5x^2 + y^2 &= 29 \quad \text{(ii)}
\end{aligned}
\(\textbf{Solution:} \)
\begin{aligned}
&x^2 + 2y^2 = 22 \quad \text{(i)} \\
&5x^2 + y^2 = 29 \quad \text{(ii)} \\
&\text{Multiplying equation (ii) by 2, we have:} \\
&10x^2 + 2y^2 = 58 \quad \text{(iii)} \\
&\text{Subtracting equation (i) from (iii):} \\
&10x^2 + 2y^2 = 58 \\
&x^2 + 2y^2 = 22 \\
&9x^2 = 36 \\
&x^2 = \frac{36}{9} = 4 \\
&\text{Taking square root, we have:} \\
&x = \pm 2 \quad \Rightarrow x = 2 \text{ or } x = -2
\end{aligned}
\(\text{Now putting these values of } x \text{ in equation (i):} \)
\begin{aligned}
&\text{When } x = -2: \\
&x^2 + 2y^2 = 22 \\
&(-2)^2 + 2y^2 = 22 \\
&4 + 2y^2 = 22 \\
&2y^2 = 22 – 4 \\
&2y^2 = 18 \\
&y^2 = \frac{18}{2} = 9 \\
&y = \pm 3 \\
&\text{When } x = 2: \\
&x^2 + 2y^2 = 22 \\
&(2)^2 + 2y^2 = 22 \\
&4 + 2y^2 = 22 \\
&2y^2 = 22 – 4 \\
&2y^2 = 18 \\
&y^2 = \frac{18}{2} = 9 \\
&y = \pm 3
\end{aligned}
\(\text{So, the solution set is: } \{(\pm 2, \pm 3)\}\)
Question 8 of Exercise 2.7 Class 10. \(4x^2 – 5y^2 = 6 \text{ and }\) \(3x^2 + y^2 = 14\)
\begin{aligned}
4x^2 – 5y^2 &= 6 \quad \text{(i)} \\
3x^2 + y^2 &= 14 \quad \text{(ii)}
\end{aligned}
\(\textbf{Solution:} \)
\begin{aligned}
&4x^2 – 5y^2 = 6 \quad \text{(i)} \\
&3x^2 + y^2 = 14 \quad \text{(ii)} \\
&\text{Multiplying equation (ii) by 5 and adding it to equation (i):} \\
&4x^2 – 5y^2 = 6 \\
&15x^2 + 5y^2 = 70 \\
&\text{Adding them gives:} \\
&19x^2 = 76 \\
&x^2 = \frac{76}{19} = 4 \quad \Rightarrow x = \pm 2
\end{aligned}
\(\text{Putting these values of } x \text{ in equation (ii):} \)
\begin{aligned}
&\text{When } x = 2: \\
&3(2)^2 + y^2 = 14 \\
&3(4) + y^2 = 14 \\
&12 + y^2 = 14 \\
&y^2 = 2 \\
&y = \pm \sqrt{2} \\
&\text{When } x = -2: \\
&3(-2)^2 + y^2 = 14 \\
&3(4) + y^2 = 14 \\
&12 + y^2 = 14 \\
&y^2 = 2 \\
&y = \pm \sqrt{2}
\end{aligned}
\(\text{So, the solution set is: }\) \(\pm 2, \pm \sqrt{2}\)
Question 9 of Exercise 2.7 Class 10. \(7x^2 – 3y^2 = 4 \text{ and } 2x^2 + 5y^2 = 7\)
\begin{aligned}
7x^2 – 3y^2 &= 4 \quad \text{(i)} \\
2x^2 + 5y^2 &= 7 \quad \text{(ii)}
\end{aligned}
\(\textbf{Solution:} \)
\begin{aligned}
&7x^2 – 3y^2 = 4 \quad \text{(i)} \\
&2x^2 + 5y^2 = 7 \quad \text{(ii)} \\
&\text{Multiply equation (i) by 5 and equation (ii) by 3 and add them:} \\
&35x^2 – 15y^2 = 20 \\
&6x^2 + 15y^2 = 21 \\
&\text{Adding them gives:} \\
&41x^2 = 41 \\
&x^2 = \frac{41}{41} = 1 \\
&x = \pm 1
\end{aligned}
\(\text{Putting these values of } x \text{ in equation (i):} \)
\begin{aligned}
&\text{When } x = 1: \\
&7(1)^2 – 3y^2 = 4 \\
&7 – 3y^2 = 4 \\
&-3y^2 = -3 \\
&y^2 = 1 \\
&y = \pm 1 \\
&\text{When } x = -1: \\
&7(-1)^2 – 3y^2 = 4 \\
&7 – 3y^2 = 4 \\
&-3y^2 = -3 \\
&y^2 = 1 \\
&y = \pm 1
\end{aligned}
\(\text{So, the solution set is: }\) \((\pm 1, \pm 1)\)
Question 10 of Exercise 2.7 Class 10. \( x^2 + 2y^2= 3 \text{ and } x^2 + 4xy – 5y^2= 0\)
\begin{aligned}
x^2 + 2y^2 &= 3 \quad \text{(i)} \\
x^2 + 4xy – 5y^2 &= 0 \quad \text{(ii)}
\end{aligned}
\(\textbf{Solution:} \)
\begin{aligned}
&x^2 + 2y^2 = 3 \quad \text{(i)} \\
&x^2 + 4xy – 5y^2 = 0 \quad \text{(ii)} \\
&\text{Factorizing equation (ii) we get:} \\
&(x + 5y)(x – y) = 0 \\
&\text{Either} \, x + 5y = 0 \, \text{or} \, x – y = 0 \\
&x = -5y \quad \text{(iii)} \quad \text{or} \quad x = y \quad \text{(iv)}
\end{aligned}
\(\text{Putting these values of } x \text{ in equation (i):} \)
\(\text{When } x = -5y: \)
\begin{aligned}
&(-5y)^2 + 2y^2 = 3 \\
&25y^2 + 2y^2 = 3 \\
&27y^2 = 3 \\
&y^2 = \frac{3}{27} = \frac{1}{9} \\
&y = \pm \frac{1}{3}
\end{aligned}
\(\text{When } x = y: \)
\begin{aligned}
&y^2 + 2y^2 = 3 \\
&3y^2 = 3 \\
&y^2 = 1 \\
&y = \pm 1
\end{aligned}
\(\text{Now putting the value of } y = \pm \frac{1}{3} \text{ in equation (iii):} \)
\begin{aligned}
\text{When } y = \frac{1}{3}: \\
&x = -5y \\
&x = -5\left(\frac{1}{3}\right) \\
&x = -\frac{5}{3} \\
\text{When } y = -\frac{1}{3}: \\
&x = -5y \\
&x = -5\left(\frac{-1}{3}\right) \\
&x = \frac{5}{3}
\end{aligned}
\(\text{Now putting the values of } y = \pm 1 \text{ in equation (iv):} \)
\begin{aligned}
&x = y \\
\text{When } y = 1, \, x = 1 \\
\text{When } y = -1, \, x = -1
\end{aligned}
\(\text{Solution set is: } \left\{(-1,-1), (1,1), \left(\frac{5}{3}, -\frac{1}{3}\right), \left(-\frac{5}{3}, \frac{1}{3}\right)\right\}\)
Question 11 of Exercise 2.7 Class 10. \( 3x^2 – y^2 = 26 \text{ and } 3x^2 – 5xy – 12y^2 = 0\)
\begin{aligned}
3x^2 – y^2 &= 26 \quad \text{(i)} \\
3x^2 – 5xy – 12y^2 &= 0 \quad \text{(ii)}
\end{aligned}
\(\textbf{Solution:} \)
\begin{aligned}
&3x^2 – y^2 = 26 \quad \text{(i)} \\
&3x^2 – 5xy – 12y^2 = 0 \quad \text{(ii)} \\
&\text{Factorizing equation (ii):} \\
&3x(x – 3y) + 4y(x – 3y) = 0 \\
&\text{Either} \, x – 3y = 0 \, \text{or} \, 3x + 4y = 0 \\
&x = 3y \quad \text{(iii)} \quad \text{or} \quad x = \frac{-4y}{3} \quad \text{(iv)}
\end{aligned}
\(\text{From equation (iii), putting the value of } x \text{ in equation (i):} \)
\begin{aligned}
3(3y)^2 – y^2 &= 26 \\
27y^2 – y^2 &= 26 \\
26y^2 = 26 \\
y^2 = 1 \\
y = \pm 1
\end{aligned}
\(\text{From equation (iv), putting the value of } x \text{ in equation (i):} \)
\begin{aligned}
&3\left(\frac{-4y}{3}\right)^2 – y^2 = 26 \\
&3\left(\frac{16y^2}{9}\right) – y^2 = 26 \\
&\frac{48y^2}{9} – y^2 = 26 \\
&\frac{48y^2 – 9y^2}{9} = 26 \\
&\frac{39y^2}{9} = 26 \\
&y^2 = \frac{26 \times 9}{39} = 6 \\
&y = \pm \sqrt{6}
\end{aligned}
\(\text{Now, using the values of } y \text{ in equations (iii) and (iv):} \)
\begin{aligned}
&\text{When } y = 1, \, x = 3y = 3 \quad \text{and} \quad x = \frac{-4(1)}{3} = -\frac{4}{3} \\
&\text{When } y = -1, \, x = 3y = -3 \quad \text{and} \quad x = \frac{-4(-1)}{3} = \frac{4}{3} \\
&\text{When } y = \sqrt{6}, \, x = \frac{-4\sqrt{6}}{3} \\
&\text{When } y = -\sqrt{6}, \, x = \frac{4\sqrt{6}}{3}
\end{aligned}
\(\text{So,solution set is: } \{(3,1), (-3,-1), \left(\frac{-4\sqrt{6}}{3}, \sqrt{6}\right), \left(\frac{4\sqrt{6}}{3}, -\sqrt{6}\right)\}\)
Question 12 of Exercise 2.7 Class 10.
\begin{aligned}
x^2 + xy &= 5 \quad \text{(i)} \\
y^2 + xy &= 3 \quad \text{(ii)}
\end{aligned}
\(\textbf{Solution:} \)
\(\text{Multiply equation (i) by 3 and equation (ii) by 5 and subtract them:} \)
\begin{aligned}
3x^2 + 3xy &= 15 \\
5xy + 5y^2 &= 15 \\
\end{aligned}
Subtracting:
\begin{aligned}
3x^2 – 2xy – 5y^2 &= 0 \\
3x^2 – 5xy + 3xy – 5y^2 &= 0 \\
x(3x – 5y) + y(3x – 5y) &= 0 \\
\text{Either} \quad 3x – 5y = 0 \quad \text{or} \quad x + y = 0 \\
3x = 5y \quad \text{or} \quad x = -y
\end{aligned}
\(\text{From equation } 3x = 5y: \)
\begin{aligned}
x &= \frac{5y}{3} \quad \text{(iii)}
\end{aligned}
\(\text{Now putting } y = -x \text{ in equation (i):} \)
\begin{aligned}
(-y)^2 + (-y)y &= 5 \\
y^2 – y^2 &= 5 \\
0 &= 5 \quad \text{(Impossible)}
\end{aligned}
\(\text{Now from equation (iii), putting } x = \frac{5y}{3} \text{ into equation (i):} \)
\begin{aligned}
\left(\frac{5y}{3}\right)^2 + \frac{5y}{3} \cdot y &= 5 \\
\frac{25y^2}{9} + \frac{5y^2}{3} &= 5 \\
\text{Multiply by 9:} \\
9 \times \frac{25y^2}{9} + 9 \times \frac{5y^2}{3} &= 9 \times 5 \\
25y^2 + 15y^2 &= 45 \\
40y^2 &= 45 \\
y^2 &= \frac{45}{40} = \frac{9}{8} \\
y &= \pm \sqrt{\frac{9}{8}} = \pm \frac{\sqrt{9}}{\sqrt{8}} = \pm \frac{3}{2\sqrt{2}}
\end{aligned}
\begin{aligned}
y &= \pm \frac{3}{2\sqrt{2}} \\
y &= \frac{3}{2\sqrt{2}} \quad \text{or} \quad y = \frac{-3}{2\sqrt{2}} \\
\end{aligned}
\(\text{Now putting the value of } y \text{ in equation (iii):}\)
\begin{aligned}
\text{When } y &= \frac{3}{2\sqrt{2}}: \\
x &= \frac{5}{3} \times \frac{3}{2\sqrt{2}} \\
x &= \frac{5}{2\sqrt{2}} \\
\left(\frac{5}{2\sqrt{2}}, \frac{3}{2\sqrt{2}}\right)
\end{aligned}
\begin{aligned}
\text{When } y &= \frac{-3}{2\sqrt{2}}: \\
x &= \frac{5}{3} \times \left(\frac{-3}{2\sqrt{2}}\right) \\
x &= \frac{-5}{2\sqrt{2}} \\
\left(\frac{-5}{2\sqrt{2}}, \frac{-3}{2\sqrt{2}}\right)
\end{aligned}
\(\text{Solution set is:} \left\{\left(\frac{5}{2\sqrt{2}}, \frac{3}{2\sqrt{2}}\right), \left(\frac{-5}{2\sqrt{2}}, \frac{-3}{2\sqrt{2}}\right)\right\}\)
Question 13 of Exercise 2.7 Class 10. \(x^2 – 2xy = 7 \text{ and } xy + 3y^2 = 2\)
\begin{aligned}
x^2 – 2xy &= 7 \quad \text{(i)} \\
xy + 3y^2 &= 2 \quad \text{(ii)}
\end{aligned}
\(\textbf{Solution:} \)
\(\text{Multiplying equation (i) by 2 and equation (ii) by 7 and subtracting them, we get:} \)
\begin{aligned}
2x^2 – 4xy &= 14 \\
7xy + 21y^2 &= 14 \\
\end{aligned}
Subtracting:
\begin{aligned}
2x^2 – 11xy – 21y^2 &= 0 \\
2x^2 – 14xy + 3xy – 21y^2 &= 0 \\
2(x – 7y)(x + 3y) &= 0 \\
\end{aligned}
\(\text{Either} \quad x – 7y = 0 \quad \text{or} \quad 2x + 3y = 0 \\
x = 7y \quad \text{(iii)} \quad \text{or} \quad x = \frac{-3}{2}y \quad \text{(iv)}\)
\(\text{From equation (iii), put } x = 7y \text{ in equation (i):} \)
\begin{aligned}
(7y)^2 – 2(7y)(y) &= 7 \\
49y^2 – 14y^2 &= 7 \\
35y^2 &= 7 \\
y^2 &= \frac{7}{35} = \frac{1}{5} \\
y &= \pm \frac{1}{\sqrt{5}} \\
\end{aligned}
\(\text{Putting these values of } y \text{ in equation (iii):} \)
\begin{aligned}
\text{When } y &= \frac{1}{\sqrt{5}}, \quad x = 7y = \frac{7}{\sqrt{5}} \\
(x, y) &= \left(\frac{7}{\sqrt{5}}, \frac{1}{\sqrt{5}}\right) \\
\text{When } y &= \frac{-1}{\sqrt{5}}, \quad x = 7y = \frac{-7}{\sqrt{5}} \\
(x, y) &= \left(\frac{-7}{\sqrt{5}}, \frac{-1}{\sqrt{5}}\right)
\end{aligned}
\(\text{From equation (iv), putting the value of } x \text{ in equation (i):} \)
\begin{aligned}
\left(\frac{-3}{2}y\right)^2 – 2\left(\frac{-3}{2}y\right)y &= 7 \\
\frac{9}{4}y^2 + 3y^2 &= 7 \\
\frac{9}{4}y^2 + \frac{12}{4}y^2 &= 7 \\
\frac{21}{4}y^2 &= 7 \\
y^2 &= \frac{28}{21} \\
y^2 &= \frac{4}{3} \\
y &= \pm \frac{2}{\sqrt{3}} \\
\end{aligned}
\(\text{Putting these values of } y \text{ in equation (iv):} \)
\begin{aligned}
\text{When } y &= \frac{2}{\sqrt{3}}, \quad x = \frac{-3}{2}y = \frac{-3}{2} \times \frac{2}{\sqrt{3}} = \frac{-3\sqrt{3}}{3} = -\sqrt{3} \\
(x, y) &= \left(-\sqrt{3}, \frac{2}{\sqrt{3}}\right) \\
\text{When } y &= \frac{-2}{\sqrt{3}}, \quad x = \frac{-3}{2}y = \frac{-3}{2} \times \frac{-2}{\sqrt{3}} = \sqrt{3} \\
(x, y) &= \left(\sqrt{3}, \frac{-2}{\sqrt{3}}\right)
\end{aligned}
\(\text{Solution set is: } \left\{\left(\frac{7}{\sqrt{5}}, \frac{1}{\sqrt{5}}\right), \left(\frac{-7}{\sqrt{5}}, \frac{-1}{\sqrt{5}}\right), \left(-\sqrt{3}, \frac{2}{\sqrt{3}}\right), \left(\sqrt{3}, \frac{-2}{\sqrt{3}}\right)\right\}\)
To further reinforce your understanding of Exercise 2.7 Class 10, we’ve included a detailed video lecture that walks you through each type of simultaneous equation, offering extra examples and tips. Watch the video below to see these methods in action, helping you gain confidence in solving both linear-quadratic and quadratic-quadratic systems. With practice, you’ll master these essential algebraic techniques and be well-prepared for more advanced problems.
For more practice, revisit Exercise 2.6, where we covered advanced concepts of quadratic equations, or move ahead to Exercise 2.8, which dives into the applications of simultaneous equations. Keep strengthening your math skills by exploring each exercise! This link has all the lectures of unit 2.
