Mix Examples - Pair of Linear Equations in Two Variables Questions in English

(A) Let the sum of the ages of the mother,father,and two daughters two years ago be $S_{-2} = 40$ years.
There are $4$ people in total.
To find the sum of their ages today (present age),we add $2$ years for each of the $4$ people:
$S_{present} = 40 + (4 \times 2) = 40 + 8 = 48$ years.
To find the sum of their ages after $3$ years from now,we add $3$ years for each of the $4$ people:
$S_{+3} = 48 + (4 \times 3) = 48 + 12 = 60$ years.
Therefore,the sum of their ages after three years will be $60$ years.

(A) $1$. Let the sum of the ages of the mother,father,and two children $x$ years ago be $y$.
$2$. The total number of people is $4$ (mother,father,and two children).
$3$. Since $x$ years have passed since that time,each person has aged by $x$ years. Therefore,the current sum of their ages is $y + (4 \times x) = y + 4x$.
$4$. We need to find the sum of their ages after $z$ years from the current time.
$5$. In $z$ years,each of the $4$ people will age by $z$ years. Thus,the total increase in the sum of their ages will be $4 \times z = 4z$.
$6$. The sum of their ages after $z$ years will be $(y + 4x) + 4z = y + 4x + 4z$.

(D) Let the length of the white stick be $W$ and the length of the black stick be $B$.
From the first part of the diagram,the total length of the combined sticks is $22 \ cm$,so $W + B = 22$.
From the second part of the diagram,the white stick is longer than the black stick by $5 \ cm$,so $W - B = 5$.
Adding the two equations: $(W + B) + (W - B) = 22 + 5$,which gives $2W = 27$.
Therefore,$W = 27 / 2 = 13.5 \ cm$.
The length of the white stick is $13.5 \ cm$.

(B) To find the correct matches,we solve each system of linear equations:
$1.$ $2x - y = 4$ and $3x - 2y = 5$. From the first,$y = 2x - 4$. Substituting into the second: $3x - 2(2x - 4) = 5 \implies 3x - 4x + 8 = 5 \implies -x = -3 \implies x = 3$. Then $y = 2(3) - 4 = 2$. So,$(1-b)$.
$2.$ $5x - y = 14$ and $4x - 3y = 9$. From the first,$y = 5x - 14$. Substituting into the second: $4x - 3(5x - 14) = 9 \implies 4x - 15x + 42 = 9 \implies -11x = -33 \implies x = 3$. Then $y = 5(3) - 14 = 1$. So,$(2-c)$.
$3.$ $4x - 3y = 5$ and $3x - y = 5$. From the second,$y = 3x - 5$. Substituting into the first: $4x - 3(3x - 5) = 5 \implies 4x - 9x + 15 = 5 \implies -5x = -10 \implies x = 2$. Then $y = 3(2) - 5 = 1$. So,$(3-d)$.
$4.$ $3x - 2y = -1$ and $2x - 5y = -8$. Multiplying the first by $5$ and the second by $2$: $15x - 10y = -5$ and $4x - 10y = -16$. Subtracting: $11x = 11 \implies x = 1$. Then $3(1) - 2y = -1 \implies -2y = -4 \implies y = 2$. So,$(4-a)$.
The correct matching is $(1-b), (2-c), (3-d), (4-a)$.

(C) From the given graph,it is observed that the two lines are parallel to each other.
Parallel lines never intersect each other at any point.
Since the solution of a pair of linear equations is the point of intersection of the lines,and these lines do not intersect,there is no common point between them.
Therefore,the pair of equations represented by these lines has no solution.

(A) Let the weight of one circle be $x$ and the weight of one square be $y$.
From the first balance,we have $3$ circles on the left and $4$ squares on the right,which are in equilibrium. Therefore,$3x = 4y$.
From the second balance,we have $1$ circle and $2$ squares on the left,and a weight of $10$ units on the right. Therefore,$x + 2y = 10$.
Thus,the pair of equations is $3x = 4y$ and $x + 2y = 10$.

(A) Let the amount of money Naresh has be $x$ and the amount of money Yogesh has be $y$.
According to the first condition,Naresh has double the money of Yogesh,so $x = 2y$.
According to the second condition,if Naresh gives ₹ $20$ to Yogesh,Naresh's money becomes $(x - 20)$ and Yogesh's money becomes $(y + 20)$.
Since they now have equal amounts,we have the equation $x - 20 = y + 20$.
Thus,the pair of equations is $x = 2y$ and $x - 20 = y + 20$.

(A) In the given figure,the two lines intersect each other at a single point.
According to the geometric interpretation of linear equations in two variables,if two lines intersect at a single point,the pair of linear equations has a unique solution (exactly one solution).
Therefore,the correct option is $A$.

(B) When two lines are coincident,every point on the line is a common solution to both equations.
Since a line consists of an infinite number of points,the pair of linear equations has infinitely many solutions.
Therefore,the correct option is $B$.

(B) For a system of linear equations $a_1x + b_1y = c_1$ and $a_2x + b_2y = c_2$:
$1$. If $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$,the system has no solution (empty set).
$2$. If $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$,the system has infinitely many solutions (infinite set).
$3$. If $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$,the system has a unique solution (singleton set).
Analysis:
$1. x+2y=3, 2x+4y=5$: Here $\frac{1}{2} = \frac{2}{4} \neq \frac{3}{5}$. This corresponds to $b$ (empty set).
$2. 2x+3y=6, x+\frac{3}{2}y=3$: Here $\frac{2}{1} = \frac{3}{1.5} = \frac{6}{3} = 2$. This corresponds to $c$ (infinite set).
$3. 3x-y=0, x-y+6=0$: Here $\frac{3}{1} \neq \frac{-1}{-1}$. This corresponds to $a$ (singleton set).
Thus,the correct match is $(1-b), (2-c), (3-a)$.