Linear Equations In Two Variables - Class 9 Mathematics - Chapter 4 - Notes, NCERT Solutions & Extra Questions
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Exercise 4.1 - Linear Equations In Two Variables | NCERT | Mathematics | Class 9
The cost of a notebook is twice the cost of a pen. Write a linear equation in two variables to represent this statement.
(Take the cost of a notebook to be ₹ $x$ and that of a pen to be ₹ $y$ ).
Given the cost of a notebook is represented by ₹ $x$ and the cost of a pen by ₹ $y$, the relationship described can be summarized as:
"The cost of a notebook is twice the cost of a pen."
Translating this into a linear equation in two variables, we get:
$$ x = 2y $$
Here, $x$ and $y$ are the variables representing the cost of a notebook and a pen, respectively.
Express the following linear equations in the form $a x+b y+c=0$ and indicate the values of $a, b$ and $c$ in each case:
(i) $2 x+3 y=9.3 \overline{5}$
(ii) $x-\frac{y}{5}-10=0$
(iii) $-2 x+3 y=6$
(iv) $x=3 y$
(v) $2 x=-5 y$
(vi) $3 x+2=0$
(vii) $y-2=0$
(viii) $5=2 x$
To express the given linear equations in the form $ax + by + c = 0$ and to indicate the values of $a, b,$ and $c$ for each case, we can rearrange each equation accordingly:
(i) $2x + 3y = 9.35\overline{5}$
This equation is already in the desired form, with $a = 2$, $b = 3$, and $c = -9.35\overline{5}$.
(ii) $x - \frac{y}{5} - 10 = 0$
Multiply everything by $5$ to clear the fraction: $5x - y - 50 = 0$, thus $a = 5$, $b = -1$, and $c = -50$.
(iii) $-2x + 3y = 6$
This equation is already in the desired form with $a = -2$, $b = 3$, and $c = -6$.
(iv) $x = 3y$
Rewriting, $x - 3y = 0$, so $a = 1$, $b = -3$, and $c = 0$.
(v) $2x = -5y$
Rewriting, $2x + 5y = 0$, thus $a = 2$, $b = 5$, and $c = 0$.
(vi) $3x + 2 = 0$
This equation is already in the form $ax + by + c = 0$ with $a = 3$, $b = 0$, and $c = 2$.
(vii) $y - 2 = 0$
Rewriting, $0x + y - 2 = 0$, thus $a = 0$, $b = 1$, and $c = -2$.
(viii) $5 = 2x$
Rewriting, $-2x + 0y + 5 = 0$, so $a = -2$, $b = 0$, and $c = 5$.
Summary of values:
- (i) $a = 2$, $b = 3$, $c = -9.35\overline{5}$
- (ii) $a = 5$, $b = -1$, $c = -50$
- (iii) $a = -2$, $b = 3$, $c = -6$
- (iv) $a = 1$, $b = -3$, $c = 0$
- (v) $a = 2$, $b = 5$, $c = 0$
- (vi) $a = 3$, $b = 0$, $c = 2$
- (vii) $a = 0$, $b = 1$, $c = -2$
- (viii) $a = -2$, $b = 0$, $c = 5$
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Ask Chatterbot AIExercise 4.2 - Linear Equations In Two Variables | NCERT | Mathematics | Class 9
Which one of the following options is true, and why? $y=3 x+5$ has
(i) a unique solution,
(ii) only two solutions,
(iii) infinitely many solutions
The equation $y = 3x + 5$ represents a straight line in the XY-plane. For any unique value of $x$, there is a unique corresponding value of $y$. This relationship between $x$ and $y$ implies that for every point $(x, y)$ that satisfies the equation, $x$ determines $y$ uniquely. Therefore, for each input value of $x$, there is exactly one output value for $y$.
Given that a straight line extends infinitely in both directions along the XY-plane, there are infinitely many points $(x, y)$ that satisfy the equation $y = 3x + 5$. Each of these points represent a unique combination of $x$ and $y$ that fits the equation, and since the line is continuous and unbounded, it crosses every possible $x$-value exactly once, yielding a corresponding $y$-value.
Therefore, the correct option is:
(iii) infinitely many solutions
The reason is that the equation depicts a linear relationship with a constant slope, indicating that for every value of $x$, there is a unique value of $y$, and since $x$ can assume any value from the set of real numbers, there are infinitely many such $(x, y)$ pairs that satisfy the equation.
Write four solutions for each of the following equations:
(i) $2 x+y=7$
(ii) $\pi x+y=9$
(iii) $x=4 y$
It seems there was an issue in the execution of my commands. Let's correct this by providing explicit solutions for each equation manually:
For equation $2x + y = 7$:
To find four solutions, we can select arbitrary values for $x$ and solve for $y$:
- Let $x = 0$, then $y = 7$.
- Let $x = 1$, then $y = 5$.
- Let $x = 2$, then $y = 3$.
- Let $x = 3$, then $y = 1$.
For equation $\pi x + y = 9$:
Similarly, we choose values for $x$ and solve for $y$:
- Let $x = 0$, then $y = 9$.
- Let $x = 1$, then $y = 9 - \pi \approx 6.14$.
- Let $x = 2$, then $y = 9 - 2\pi \approx 3.28$.
- Let $x = 3$, then $y = 9 - 3\pi \approx 0.42$.
For equation $x = 4y$:
For this, we select values for $y$ and solve for $x$:
- Let $y = 0$, then $x = 0$.
- Let $y = 1$, then $x = 4$.
- Let $y = 2$, then $x = 8$.
- Let $y = -1$, then $x = -4$.
These selections give us four explicit solutions for each equation.
Check which of the following are solutions of the equation $x-2 y=4$ and which are not:
(i) $(0,2)$
(ii) $(2,0)$
(iii) $(4,0)$
(iv) $(\sqrt{2}, 4 \sqrt{2})$
(v) $(1,1)$
To check if each given ordered pair is a solution to the equation $x - 2y = 4$, we simply substitute the values of $x$ and $y$ from each pair into the equation and see if it holds true. Let's check each one.
-
For $(0,2)$: Substituting $x = 0$ and $y = 2$.
-
For $(2,0)$: Substituting $x = 2$ and $y = 0$.
-
For $(4,0)$: Substituting $x = 4$ and $y = 0$.
-
For $(\sqrt{2}, 4 \sqrt{2})$: Substituting $x = \sqrt{2}$ and $y = 4 \sqrt{2}$.
-
For $(1,1)$: Substituting $x = 1$ and $y = 1$.
Let me calculate each and provide the results. Let's review which of the given ordered pairs are solutions to the equation $x - 2y = 4$:
- For $(0,2)$, the equation yields a result of $-4$, so it is not a solution.
- For $(2,0)$, the equation yields a result of $2$, so it is not a solution.
- For $(4,0)$, the equation yields a result of $4$, so it is a solution.
- For $(\sqrt{2}, 4 \sqrt{2})$, the equation yields a result of approximately $-9.90$, so it is not a solution.
- For $(1,1)$, the equation yields a result of $-1$, so it is not a solution.
Therefore, only the pair $(4,0)$ is a solution to the equation $x - 2y = 4$.
Find the value of $k$, if $x=2, y=1$ is a solution of the equation $2 x+3 y=k$.
The value of $k$, if $x=2, y=1$ is a solution of the equation $2 x+3 y=k$, is $7$.
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Ask Chatterbot AIExtra Questions - Linear Equations In Two Variables | NCERT | Mathematics | Class 9
The radius of a circle is proportional to its diameter. Find the constant of proportionality.
A) $\frac{1}{2}$
B) 5
C) $2 \pi$
D) $\pi$
The relationship between the radius and the diameter of a circle is governed by the equation:
$$ \text{Radius} = \frac{1}{2} \times \text{Diameter} $$
Here, the constant of proportionality between the radius and the diameter is $\frac{1}{2}$. This constant represents the fraction of the diameter that equals the radius.
Thus, the correct answer is:
A) $\frac{1}{2}$
Give two solutions for $3y+4=0$.
Step-by-Step Solution:
Addressing as a Single Variable Equation:
The equation provided is $3y+4=0$. Solving for $y$, we can manipulate the equation as follows: $$ \begin{align*} 3y + 4 &= 0 \ 3y &= -4 \ y &= -\frac{4}{3} \end{align*} $$ As a linear equation in one variable (y), this has the unique solution $y = -\frac{4}{3}$.Treating it as a Two Variable Equation:
For a linear equation representing a line in a two-dimensional space, we can introduce another variable (x), which can take any value, giving us infinitely many solutions in the form ((x, y)). Keeping $y = -\frac{4}{3}$ as derived, we see that:Example 1:If $x = 1$, the solution formulates as: $$ \begin{align*} 3y + 0 \cdot 1 + 4 &= 0 \ y &= -\frac{4}{3} \end{align*} $$ Therefore, one solution is $(1, -\frac{4}{3})$.
Example 2:If $x = 2$, the solution remains the same in $y$ as: $$ \begin{align*} 3y + 0 \cdot 2 + 4 &= 0 \ y &= -\frac{4}{3} \end{align*} $$ Hence, another solution is $(2, -\frac{4}{3})$.
Conclusion:
The equation $3y+4=0$, when viewed in one variable, uniquely determines $y = -\frac{4}{3}$.
When extended to two variables where $x$ can be any real number, it describes a line with infinitely many solutions, exemplified by $(1, -\frac{4}{3})$ and $(2, -\frac{4}{3})$ among others.
There are 7 observations in a set of data and their mean is 11. If each observation is multiplied by 2, find the new mean.
Formula Used for Mean (1 Mark): The mean of a set of observations is calculated using the formula: $$ \text{Mean} = \frac{\text{Sum of observations}}{\text{Number of observations}} $$
Given:
Mean = 11
Number of observations = 7
From the formula, substituting the given values: $$ 11 = \frac{\text{Sum of observations}}{7} $$ This equation can be rearranged to find the sum of the observations: $$ \text{Sum of observations} = 11 \times 7 = 77 $$
Modification of Observations: If each observation is multiplied by 2, then the new sum of observations is: $$ \text{New Sum of observations} = 77 \times 2 = 154 $$
Calculation of New Mean: The new mean is calculated by dividing the new sum of observations by the number of observations: $$ \text{New Mean} = \frac{154}{7} = 22 $$
Therefore, the new mean, after each observation is multiplied by 2, is 22.
The graph of the linear equation $2x + 3y = 6$ is a line which meets the $x$-axis at the point:
A) $(0, 2)$
B) $(2, 0)$
C) $(3, 0)$
D) $(0, 3)$
Solution
The correct answer is Option C $(3,0)$.
To determine where the line $2x + 3y = 6$ intersects the $x$-axis, set $y = 0$ (since all points on the $x$-axis have zero $y$-coordinate):
$$ 2x + 3(0) = 6 \rightarrow 2x = 6 \rightarrow x = 3 $$
Thus, the point of intersection with the $x$-axis is $(3, 0)$. We can validate this by plugging these coordinates back into the original equation, confirming the left and right-hand sides are equal. Hence, Option C is the correct choice.
Find the equation of the line which passes through the point $(3,4)$ and the sum of its intercepts on the axes is 14.
A) $2x + y - 7 = 0$
B) $x + y - 7 = 0$
C) $x - y - 7 = 0$
D) $x + y + 7 = 0
The equation of a line passing through a point $(3,4)$ and having the sum of its x-axis and y-axis intercepts equal to 14 can be derived using the intercept form and substituting the given conditions.
Given:
- The sum of the intercepts $a+b = 14$, where $a$ is the x-intercept and $b$ is the y-intercept.
- Points on the axes can be (0, b) and (a, 0). Given the sum, these points are:
- $(0, b) = (0, 14-a)$
- $(a, 0)$
The equation of the line represented through points $A$ and $B$ is: $$ \begin{array}{l} y - (14-a) = \frac{0 - (14-a)}{a - 0}(x - 0) \ \Rightarrow y - 14 + a = \frac{a - 14}{a} x, \quad \text{(i)} \end{array} $$
This equation must pass through point $(3, 4)$. Thus, by substituting $(3, 4)$ into equation (i): $$ 4 - 14 + a = \frac{a - 14}{a} \cdot 3 \ \Rightarrow a - 10 = \frac{3a - 42}{a} \ \Rightarrow a^2 - 10a - 3a + 42 = 0 \ \Rightarrow a^2 - 13a + 42 = 0 \ \Rightarrow (a-7)(a-6) = 0 $$
Solving the quadratic, we obtain $a = 7$ or $a = 6$. We need to find the correct equation matching one of the choices:
-
For $a = 6$, substitute into equation (i): $$ y - 14 + 6 = \frac{6 - 14}{6}x \ \Rightarrow y - 8 = -\frac{4}{3}x \ \Rightarrow 3y - 24 = -4x \ \Rightarrow 4x + 3y - 24 = 0 $$ This is not listed among the options presented.
-
For $a = 7$, substitute into equation (i): $$ y - 14 + 7 = \frac{7 - 14}{7}x \ \Rightarrow y - 7 = -x \ \Rightarrow x + y - 7 = 0 $$ This matches option B) $x + y - 7 = 0$.
Therefore, the correct equation is $x + y - 7 = 0$ (Option B).
If the mean of the following distribution is 2.6, then the value of $y$ is \begin{tabular}{|l|l|l|l|l|l|} \hline Variable $(x)$ & 1 & 2 & 3 & 4 & 5 \ \hline Frequency & 4 & 5 & $y$ & 1 & 2 \ \hline \end{tabular}
A) 3
B) 8
C) 13
D) $24$
To find the value of ( y ) in the given distribution, where the mean is ( 2.6 ), we proceed with the following calculations:
-
Calculate the total frequency, ( n ) $$ n = \sum f_i = 4 + 5 + y + 1 + 2 = 12 + y $$
-
Calculate the sum of the products of variables and their frequencies, ( \sum f_i x_i ) $$ \sum f_i x_i = 1 \times 4 + 2 \times 5 + 3 \times y + 4 \times 1 + 5 \times 2 = 4 + 10 + 3y + 4 + 10 = 28 + 3y $$
-
Set up the equation for the mean and solve for ( y ) Using the formula for the mean: $$ \text{Mean} = \frac{\sum f_i x_i}{n} \Rightarrow 2.6 = \frac{28 + 3y}{12 + y} $$ Cross-multiplying yields: $$ 26(12 + y) = 280 + 30y $$ Simplifying this, we get: $$ 312 + 26y = 280 + 30y \Rightarrow 4y = 32 \Rightarrow y = 8 $$
Thus, the value of ( y ) is 8, which corresponds to option B. The calculations confirm that the correct choice is indeed B) 8.
Divide $2x^{3}y - 6x^{2}y^{3} + 6x^{2}y^{2}$ by $2x^{2}y$.
A) $x - 3y^{2} + 2y$ B) $x - 2y^{2} + 3y$ C) $x - 3y^{2} + 3y$ D) $x - 3y^{2} - 3y$
Given the problem, we need to divide each term of the polynomial $$ 2x^3y - 6x^2y^3 + 6x^2y^2 $$ by $$ 2x^2y $$. We proceed by handling each term separately:
-
For the first term: $$ \frac{2x^3y}{2x^2y} = x $$ (since $2x^3y$ divided by $2x^2y$ reduces to $x$ as $2x^3y / 2x^2y = x^1y^0 = x$).
-
For the second term: $$ \frac{-6x^2y^3}{2x^2y} = -3y^2 $$ (since $-6x^2y^3$ divided by $2x^2y$ simplifies to $-3y^2$ as $-6x^2y^3 / 2x^2y = -3y^2$).
-
For the third term: $$ \frac{6x^2y^2}{2x^2y} = 3y $$ (since $6x^2y^2$ divided by $2x^2y$ simplifies to $3y$ as $6x^2y^2 / 2x^2y = 3y$).
Combining the results of these divisions, we have: $$ x - 3y^2 + 3y $$.
Thus, the correct answer to the given problem is: $x - 3y^2 + 3y$, which corresponds to Option C.
Slope of the line joining the points $(2,1)$ and $(6,7)$ is
A) $\frac{3}{2}$
B) $\frac{2}{3}$
C) 1
D) 0.0
The correct answer is A) $\frac{3}{2}$.
To find the slope of the line joining the points $(2,1)$ and $(6,7)$, apply the slope formula: $$ \text{slope} = \frac{y_2 - y_1}{x_2 - x_1} $$
Plugging in the coordinates $(x_1, y_1) = (2, 1)$ and $(x_2, y_2) = (6, 7)$: $$ \text{slope} = \frac{7 - 1}{6 - 2} = \frac{6}{4} = \frac{3}{2} $$
Therefore, the slope of the line is $\frac{3}{2}$.
The equation of a line parallel to the $y$-axis and passing through $(-3,-2)$ is
(A) $x - 3 = 0$ (B) $x + 3 = 0$ (C) $y + 2 = 0$ (D) $y - 2 = 0$
Solution
The correct answer is (B) $x + 3 = 0$.
A line parallel to the $y$-axis takes the form $x = a$, where $a$ is the x-coordinate at which the line intersects or passes through. Given the point $(-3,-2)$, the line must pass through $x = -3$. Thus, the equation can be reformulated as:
$$ x + 3 = 0 $$
This confirms that option (B) is the correct choice, as it represents the line passing through the specified point with the required condition of being parallel to the $y$-axis.
The slope of the line passing through points $(3,2)$ and $(2,5)$ is:
A) $\frac{1}{3}$ B) -3 C) $-\frac{1}{3}$ D) 3
Solution
The slope of a line passing through two points, $(x_1, y_1)$ and $(x_2, y_2)$, can be calculated using the formula:
$$ m = \frac{y_2 - y_1}{x_2 - x_1} $$
By substituting the given points $(3,2)$ as $(x_1, y_1)$ and $(2,5)$ as $(x_2, y_2)$ into the formula, we get:
$$ m = \frac{5 - 2}{2 - 3} = \frac{3}{-1} = -3 $$
Thus, the correct answer is B) -3.
The slope of a line passing through points $(1,2)$ and $(10,20)$ is:
To calculate the slope of a line that passes through two points $(x_1, y_1)$ and $(x_2, y_2)$, we use the formula: $$ \text{slope} = \frac{y_2 - y_1}{x_2 - x_1} $$ Given points are $(1, 2)$ and $(10, 20)$. Plugging these values into the formula, we get: $$ \text{slope} = \frac{20 - 2}{10 - 1} = \frac{18}{9} = 2 $$ Thus, the slope of the line passing through the points $(1,2)$ and $(10,20)$ is $2$.
If $x^{2} - y^{2} + 4x - 6y + k$ is resolvable into two linear factors, then $k =$
A. -1
B. 4
C. 6
D. $-5$
To find the value of $ k $ such that the expression $ x^{2} - y^{2} + 4x - 6y + k $ can be resolved into two linear factors, we need to manipulate and factorize the given polynomial.
Given: $$ x^{2} - y^{2} + 4x - 6y + k $$
First, let's rearrange and group the terms for completing the square for both $ x $ and $ y $:
Group the $ x $ terms together and the $ y $ terms together: $$ (x^2 + 4x) - (y^2 - 6y) + k $$
Complete the square for $ x $:
$ x^2 + 4x $ can be written as $ (x+2)^2 - 4 $.
Complete the square for $ y $:
$ y^2 - 6y $ can be written as $ (y-3)^2 - 9 $.
Now, substituting these back into the original equation:
$$ (x+2)^2 - 4 - ( (y-3)^2 - 9 ) + k $$
Simplify further:
$$ (x+2)^2 - 4 - (y-3)^2 + 9 + k $$
$$ (x+2)^2 - (y-3)^2 + 5 + k $$
For the polynomial to factor into two linear factors, the entire expression must be a difference of squares:
$$ (x+2)^2 - (y-3)^2 = [ (x+2) + (y-3) ][ (x+2) - (y-3) ] $$
Thus, we need:
$$ k + 5 = 0 \implies k = -5 $$
Therefore, the value of $ k $ that allows $ x^{2} - y^{2} + 4x - 6y + k $ to be resolved into two linear factors is $ \boxed{-5} $.
Final Answer: D. (-5)
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