Algebra Chapter 3 Linear Simultaneous Equations
Linear equations:
⇒ You have learned from your Algebra Chapter 3 Linear Simultaneous Equations about the equations in one and two variables, and simultaneous linear equations in one and two variables.
⇒ You have learned how graphs of these linear equations are drawn on graph papers and also how they are solved with the help of a graph.
⇒ The general standard form of two simultaneous linear equations in two variables are:
⇒ a1x + b1y + c1 = 0 where a1, b1, c1 are constants and both a1 and b1, are not zero simultaneously.
⇒ a2x+b2y+c2 =0, where a2, b2, c2 are constants and both a2, b2 are not zero simultaneously.
⇔ You can solve two equations by two methods
Read and Learn More WBBSE Solutions For Class 9 Maths
By drawing graphs:
- By finding the relations among the ratios of the coefficients of the same variables and of the constant terms of the two equations. By drawing graph if those two straight lines intersect to each other, then the two simultaneous equations are solvable and there is one and only one solution set of these two equations.
- If two straight lines coincide then the given two simultaneous linear equations are solvable and there are infinitely many solutions sets of these two equations.
- If two straight lines are parallel to each other then the given two simultaneous linear equations are not solvable. There is no solution set of these equations.
By finding the relations among the ratios between the coefficients of the same variable and the constant terms of the two equations:
- If \(\frac{a_1}{a_2} \neq \frac{b_1}{b_2}\), then the given two simultaneous linear equations are solvable and there is one and only one set of solutions in this case.
- If \(\frac{a_1}{a_2}=\frac{b_1}{b_2} \neq \frac{c_1}{c_2}\), then the given two simultaneous linear equations are solvable and there are infinitely many sets of solutions in this case.
- \(\frac{a_1}{a_2}=\frac{b_1}{b_2} \neq \frac{c_1}{c_2}\), then the given simultaneous linear equations are not solvable and there is
no set of solutions in this case.
Now we can say that,
- If \(\frac{a_1}{a_2} \neq \frac{b_1}{b_2}\), then the graph of two given simultaneous equations intersect each other.
- If \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\), then the graph of the given two simultaneous linear equations coincide.
- If \(\frac{a_1}{a_2}=\frac{b_1}{b_2} \neq \frac{c_1}{c_2}\), then the graph of the given two simultaneous linear equations are parallel to each other.
Algebra Chapter 3 Linear Simultaneous Equations True Or False
Example 1. The equations 2x + 3y -7 = 0 and 3x + 2y – 8 = 0 are solvable
Solution: The statement is True.
Example 2. The equations x + 5y = 7, and x + 5y = 20 cannot be solved.
Solution: The statement is True.
\(\frac{1}{1}=\frac{5}{5} \neq \frac{7}{20}\)
Example 3. The equations 5x + 3y = 11, and 2x – 7y = -12 have an infinite no. of solutions.
Solution: The statement is False.
\(\frac{5}{2} \neq \frac{3}{-7}\)
Example 4. Two equations x + y = 5 and 2x + 2y = 10 are overlapping.
Solution: The statement is True.
Example 5. The equations 3x – 4y = 1 and 9x + 8y = 2 have one and only one solution.
Solution: The statement is True.
Example 6. The value of r for which rx + 2y = 5 and (r + 1) x + 3y = 2 will have no solution is 3.
Solution: The statement is False.
Example 7. x + y = 20 and 10x + 5y = 140 are solvable and there is only one common solution.
Solution: The statement is True.
Example 8. There are 3 methods to solve two linear simultaneous equations of two variables.
Solution: The statement is False. The correct answer is Four.
Example 9. We can substitute the value of one variable by another variable in the elimination method.
Solution: The statement is True. (substitution method)
Example 10. From the equations \(a_1 x+b_1 y+c_1=0 \text { and } a_2 x+b^2 y+c_2=0\) we get by method of cross multiplication, \(\frac{x}{b_1 c_2-b_2 c_1}=\frac{y}{c_2 a_2-c_2 a_1}=\frac{1}{a_1 b_2-a_2 b_1}\)
Solution: The statement is True.
Algebra Chapter 3 Linear Simultaneous Equations Fill In The Blanks
Example 1. If the equation rx + 2y = 5 and (r – 5) x + 3y = 2 have no solution then r = _______
Solution: 10.
Example 2. If the equation rx + 2y = 5 and (r + 1)x + 3y = 2 have no solution then r = _______
Solution: 2
Example 3. If the simultaneous linear equations 3x + 4y = 18 and kx – 4y = 180 have no solution then the value of k is _______
Solution: -3.
Example 4. The simultaneous linear equations in two variables will be inconsistent of their graph is _______
Solution: parallel.
Example 5. The straight lines x + y = p and \(\frac{1}{2} x+\frac{1}{2} y=\frac{1}{2} p\) are _______
Solution: parallel.
Algebra Chapter 3 Linear Simultaneous Equations Short Answer Type Questions
Example 1. If x = 3t and y = \(\frac{2t}{3}\) – 1 then for what velue of 1, x = 3y.
Solution: x = 3y
\(3 t=3\left(\frac{2 t}{3}-1\right)\)
⇒ or, 3t = 2t – 3
⇒ or, t = -3
Example 2. For what value of k two equation 2x + 5y = 8 and 2x – ky = 3 will have no solution?
Solution: \(\frac{2}{2}=\frac{5}{-k} \neq \frac{8}{3}\)
or, \(\frac{5}{-k}\) = 1 or, k = -5
Example 3. If x and y are real numbers and (x – 5)2 + (x – y)2 = 0, then what are the values of x and y?
Solution: If sum of the squares is zero, then each is zero.
∴ x – 5 = 0, x = 5 and x – y = 0, x = y = 5
Example 4. If x2 + y2 – 2x + 4y =-5, then find the values of x and y.
Solution: (x – y)2 + (y + 2)2 = 0
If the sum of the squares is zero, then each term is zero.
∴ x – y = 0, x = y and y + 2 = 0
⇒ or, y = -2
∴ x = y = 2.
Example 5. For what values of r, the two equations rx – 3y – 1 = 0 and (4 – r) x – y + 1 = 0 would have no solution?
Solution: For no solution \(\frac{r}{4-r}=\frac{-3}{-1} \neq \frac{1}{-1}\)
∴ -r = -12 + 3r
⇒ 4r = 12
⇒ or, r = 3
Example 6. Let us write the equation a1x + b1y + c1 = 0 in the form y = mx + c where m and c are constants.
Solution: a1x+b1y+ c1 = 0,
\(y_1=\frac{-a_1 x-c_1}{b_1} \quad \text { or, } \quad y_1=-\frac{a_1}{b_1} x-\frac{c_1}{b_1}\)
Example 7. For what value of k, the two equations kx – 21y + 15 = 0 and 8x – 7y = 0 have only one solution.
Solution: For only one solution, \(\frac{k}{8} \neq \frac{-21}{-7}\)
or, \(\frac{k}{8} \neq 3\)
or, k ≠ 24
Example 8. For what values of a and b, the two equations 5x + 8y = 7 and (a + b) x + (a – b) y = (2a + b + 1) have infinite number of solutions?
Solution: For infinite number of solution, \(\frac{5}{a+b}=\frac{8}{a-b}=\frac{7}{2 a+b+1}\)
⇒ or, 5a – 5b = 8a + 8b
⇒ or, 3a = -13b and, 16a + 8b + 8 = 7a – 7b
⇒ or, 9a + 15b = -8 or, 3·3a + 15b = -8
⇒ or, -24b = -8
⇒ b = \(\frac{1}{3}\)
∴ a = \(\frac{-13}{3} \times \frac{1}{3}=\frac{-13}{9}\)
Example 9. For what value of x will the two expressions \(\frac{3 x-1}{2} \text { and } \frac{2 x+6}{3}\) have the same value?
Solution: By condition, \(\frac{3 x-1}{2}\) = \(\frac{2 x+6}{3}\)
⇒ or, 3(3x – 1)= 2 (2x + 6)
⇒ or, 9x – 3 = 4x + 12
⇒ or, 5x 15, x = 3.
Example 10. For what value of p, the equation 3(x + 5)= 2p(x + 10) cannot have any solution?
Solution: 3(x + 5)= 2p(x + 10).
⇒ or, 3x + 15 = 2px + 20p
⇒ or, x (3 – 2p) = 20p – 15
⇒ or, x = \(=\frac{20 p-15}{3-2 p}\)
⇒ Clearly, the value of x cannot be determined when 3 – 2p = 0, or, p = \(\frac{3}{2}\)
⇒ When p = \(\frac{3}{2}\), the given equation cannot have any solution.
Example 11. If \(m=\sqrt{\frac{n}{x+b}}\) then express n interms of m and b.
Solution: \(m=\sqrt{\frac{n}{x+b}}\), \(m^2=\frac{n}{x+b}\)
or, \(\left(1-m^2\right) n=m^2 b\), \(n=\frac{m^2 b}{1-m^2}\)
Example 12. If x, y, z are real numbers and (x – 5)2 + (x − y)2 + (z + 4)2 = 0, find x, y, z
Solution: If the sum of three squares is zero, then each term is zero, x – 5 = 0, x = 5
∴ x – y = 0, x = y = 5, z + 4 = 0, z = -4
∴ x = y = 5, z = -4