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

(C) Given equations are $3x - 2y = 2$ and $6x - 4y = 4$.
Comparing these with the standard form $a_1x + b_1y = c_1$ and $a_2x + b_2y = c_2$:
For the first equation: $a_1 = 3, b_1 = -2, c_1 = 2$.
For the second equation: $a_2 = 6, b_2 = -4, c_2 = 4$.
Now,find the ratios:
$\frac{a_1}{a_2} = \frac{3}{6} = \frac{1}{2}$
$\frac{b_1}{b_2} = \frac{-2}{-4} = \frac{1}{2}$
$\frac{c_1}{c_2} = \frac{2}{4} = \frac{1}{2}$
Since $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} = \frac{1}{2}$,the lines are coincident lines.

(C) Two lines are parallel if their slopes are equal.
The given line is $x+3y=5$,which can be written in the slope-intercept form $y = mx + c$ as $3y = -x + 5$,or $y = -\frac{1}{3}x + \frac{5}{3}$.
The slope of this line is $m = -\frac{1}{3}$.
$A$ line parallel to this will have the same slope $m = -\frac{1}{3}$ and will be of the form $x+3y=k$ for some constant $k$.
Comparing the given options,the equation $x+3y=10$ has the same coefficients for $x$ and $y$ as the original line,meaning it has the same slope.
Therefore,$x+3y=10$ is parallel to $x+3y=5$.

(A) To find the point where the line intersects the $X$-axis,we set the $y$-coordinate to $0$.
Substituting $y = 0$ into the equation $2x - y = 6$:
$2x - 0 = 6$
$2x = 6$
$x = 3$
Therefore,the point of intersection is $(3, 0)$.

(C) To find the point where the line intersects the $Y$-axis,we set the $x$-coordinate to $0$.
Substituting $x=0$ into the equation $x-2y=2$:
$0-2y=2$
$-2y=2$
$y=-1$
Therefore,the line intersects the $Y$-axis at the point $(0,-1)$.

(D) two-digit number is formed by the sum of the value of its tens digit and its units digit.
Given that the digit at the tens place is $y$,its value is $10 \times y = 10y$.
Given that the digit at the units place is $x$,its value is $1 \times x = x$.
Therefore,the number is the sum of these values,which is $10y + x$.

(B) two-digit number is represented as $10 \times (\text{tens digit}) + (\text{units digit})$.
Given that the tens digit is $y$ and the units digit is $x$,the original number is $10y + x$.
When the digits are interchanged,the new tens digit becomes $x$ and the new units digit becomes $y$.
Therefore,the new number obtained by interchanging the digits is $10x + y$.

(C) Let the two-digit number be represented as $10x + y$,where $x$ is the tens digit and $y$ is the units digit.
When the digits are interchanged,the new number becomes $10y + x$.
The sum of these two numbers is $(10x + y) + (10y + x) = 11x + 11y$.
Factoring out $11$,we get $11(x + y)$.
Since the sum is $11(x + y)$,it is always divisible by $11$.

(C) Let the digit at the ten's place be $x$ $(x \neq 0)$ and the digit at the unit's place be $y$.
According to the problem,the sum of the digits is equal to the difference of the digits.
So,$x + y = x - y$.
Subtracting $x$ from both sides,we get $y = -y$.
Adding $y$ to both sides,we get $2y = 0$.
Therefore,$y = 0$.
Thus,the digit at the unit's place must be $0$.

(D) Let the two-digit number be represented as $10x + y$,where $x$ is the tens digit and $y$ is the units digit.
According to the problem,the sum of the digits is equal to the product of the digits:
$x + y = xy$
Rearranging the equation to solve for $y$:
$x = xy - y$
$x = y(x - 1)$
$y = \frac{x}{x - 1}$
Since $x$ and $y$ must be integers between $1$ and $9$ (with $x \neq 0$ for a two-digit number):
If $x = 2$,then $y = \frac{2}{2 - 1} = 2$.
The number is $10(2) + 2 = 22$.
Check: Sum of digits $= 2 + 2 = 4$; Product of digits $= 2 \times 2 = 4$. Both are equal.
Thus,the number is $22$.

(D) Let the two numbers be $x$ and $y,$ where $x > y.$
According to the problem:
$x + y = 15$ (Equation $1$)
$x - y = 1$ (Equation $2$)
Adding Equation $1$ and Equation $2$:
$(x + y) + (x - y) = 15 + 1$
$2x = 16$
$x = 8$
Substituting $x = 8$ into Equation $1$:
$8 + y = 15$
$y = 15 - 8$
$y = 7$
The two numbers are $8$ and $7.$ The smaller number is $7.$

(D) Given that $x$ is the larger number and $y$ is the smaller number.
Two times the larger number is $2x$.
Five times the smaller number is $5y$.
According to the problem,two times the larger number is $4$ more than five times the smaller number.
Therefore,the equation is $2x = 5y + 4$,which can be rewritten as $2x - 5y = 4$.

(C) Let the larger number be $x$ and the smaller number be $y$.
According to the problem,the difference between the two numbers is $5$.
Therefore,the equation is $x - y = 5$.
To find the smaller number $y$ in terms of $x$,we rearrange the equation:
$x - 5 = y$
Thus,$y = x - 5$.

(D) Let the two numbers be $x$ and $y$.
According to the problem,the sum of the two numbers is $8$.
Therefore,$x + y = 8$.
To find the larger number $x$ in terms of the smaller number $y$,we rearrange the equation:
$x = 8 - y$.
Thus,the correct option is $D$.

(C) Let the present age of the mother be $M$ and the present ages of the two daughters be $D_1$ and $D_2$.
Four years ago,the sum of their ages was $(M-4) + (D_1-4) + (D_2-4) = x$.
This simplifies to $M + D_1 + D_2 - 12 = x$,which means the current sum of their ages is $M + D_1 + D_2 = x + 12$.
We need to find the sum of their ages after two years,which is $(M+2) + (D_1+2) + (D_2+2)$.
This expression equals $(M + D_1 + D_2) + 6$.
Substituting the current sum $(x + 12)$ into this expression,we get $(x + 12) + 6 = x + 18$.
Therefore,the sum of their ages after two years will be $x + 18$ years.

(C) Let the present age of Ramesh be $x$ years.
Mahesh is $5$ years elder than Ramesh,so the age of Mahesh is $(x + 5)$ years.
Mahesh is $3$ years younger than Suresh,which means Suresh is $3$ years older than Mahesh.
Therefore,the age of Suresh = (Age of Mahesh) $+ 3$.
Age of Suresh = $(x + 5) + 3 = x + 8$ years.
Thus,the present age of Suresh is $(x + 8)$ years.

(C) two-digit number can be expressed as $10 \times (\text{digit at ten's place}) + (\text{digit at unit's place})$.
Given that the digit at the ten's place is $2x$ and the digit at the unit's place is $3x$.
Substituting these values into the expression:
Number $= 10(2x) + 3x$
Number $= 20x + 3x$
Number $= 23x$.
Therefore,the correct option is $C$.

(D) The digit at the unit's place is $(x+1)$.
The digit at the ten's place is $(x+2)$.
$A$ two-digit number is represented as $10 \times (\text{digit at ten's place}) + (\text{digit at unit's place})$.
Therefore,the number $= 10(x+2) + (x+1)$.
$= 10x + 20 + x + 1$.
$= 11x + 21$.

(C) Let the present ages of the father and the two sons be $F$,$S_1$,and $S_2$ respectively.
Five years ago,the sum of their ages was $(F-5) + (S_1-5) + (S_2-5) = x$.
This simplifies to $F + S_1 + S_2 - 15 = x$,which means the present sum of their ages is $F + S_1 + S_2 = x + 15$.
We need to find the sum of their ages after five years from now.
The sum after five years will be $(F+5) + (S_1+5) + (S_2+5) = (F + S_1 + S_2) + 15$.
Substituting the present sum $(x + 15)$ into this expression:
Sum $= (x + 15) + 15 = x + 30$.
Therefore,the sum of their ages after five years will be $x + 30$ years.

(A) Let the digit at the ten's place be $x$ and the digit at the unit's place be $y$.
According to the problem,the sum of the digits is $6$,so $x + y = 6$.
Also,the digit at the unit's place is twice the digit at the ten's place,so $y = 2x$.
Substituting $y = 2x$ into the first equation: $x + 2x = 6$.
$3x = 6$,which gives $x = 2$.
Now,find $y$: $y = 2(2) = 4$.
The number is formed by $10x + y$,which is $10(2) + 4 = 24$.

(A) system of linear equations in two variables $x$ and $y$ is represented as a coordinate pair $(x, y)$.
Given the equations $x = 4$ and $y = 5$,the value of $x$ is fixed at $4$ and the value of $y$ is fixed at $5$.
Therefore,the unique solution point where these two lines intersect is $(4, 5)$.
Thus,the solution set is ${(4, 5)}$.

(D) two-digit number can be expressed as $10 \times (\text{digit at ten's place}) + (\text{digit at unit's place})$.
Given that the digit at the unit's place is $x$ and the digit at the ten's place is $2x$.
Substituting these values into the expression:
Number $= 10(2x) + x$
Number $= 20x + x$
Number $= 21x$.
Therefore, the correct option is $D$.

(D) Given equations are $x+y-1=0$ and $2x+2y-2=0$.
Comparing with $a_1x+b_1y+c_1=0$ and $a_2x+b_2y+c_2=0$,we have $a_1=1, b_1=1, c_1=-1$ and $a_2=2, b_2=2, c_2=-2$.
Calculating the ratios: $\frac{a_1}{a_2} = \frac{1}{2}$,$\frac{b_1}{b_2} = \frac{1}{2}$,and $\frac{c_1}{c_2} = \frac{-1}{-2} = \frac{1}{2}$.
Since $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$,the lines are coincident.
Coincident lines have infinitely many solutions,meaning the solution set is an infinite set.

(D) Given equations are:
$3x + 3y - 3 = 0$ ... $(1)$
$5x + 5y - 5 = 0$ ... $(2)$
Divide equation $(1)$ by $3$: $x + y - 1 = 0 \implies x + y = 1$.
Divide equation $(2)$ by $5$: $x + y - 1 = 0 \implies x + y = 1$.
Since both equations represent the same line $(x + y = 1)$,they are coincident lines.
Coincident lines have infinitely many solutions.
Therefore,the solution set is an infinite set.

(D) Let $u = \frac{1}{x}$ and $v = \frac{1}{y}$.
The given equations become:
$5u - 3v = 9$ ---$(1)$
$3u - 5v = 7$ ---$(2)$
Subtracting equation $(2)$ from equation $(1)$:
$(5u - 3v) - (3u - 5v) = 9 - 7$
$2u + 2v = 2$
$u + v = 1$ ---$(3)$
Adding equation $(1)$ and equation $(2)$:
$(5u - 3v) + (3u - 5v) = 9 + 7$
$8u - 8v = 16$
$u - v = 2$ ---$(4)$
Since $u = \frac{1}{x}$ and $v = \frac{1}{y}$,the value of $\frac{1}{x} - \frac{1}{y}$ is $u - v = 2$.

(A) Let the current ages of the five persons be $x_1, x_2, x_3, x_4,$ and $x_5$.
Five years ago,the sum of their ages was $(x_1 - 5) + (x_2 - 5) + (x_3 - 5) + (x_4 - 5) + (x_5 - 5) = 50$.
This simplifies to $(x_1 + x_2 + x_3 + x_4 + x_5) - 25 = 50$,which means the current sum of their ages is $x_1 + x_2 + x_3 + x_4 + x_5 = 75$.
After five years,the sum of their ages will be $(x_1 + 5) + (x_2 + 5) + (x_3 + 5) + (x_4 + 5) + (x_5 + 5)$.
This is equal to $(x_1 + x_2 + x_3 + x_4 + x_5) + 25$.
Substituting the current sum,we get $75 + 25 = 100$ years.

(D) Let the present ages of the four persons be $x_1, x_2, x_3,$ and $x_4$.
Five years ago,their ages were $(x_1 - 5), (x_2 - 5), (x_3 - 5),$ and $(x_4 - 5)$.
According to the problem,the sum of their ages five years ago was $60$ years:
$(x_1 - 5) + (x_2 - 5) + (x_3 - 5) + (x_4 - 5) = 60$
$(x_1 + x_2 + x_3 + x_4) - 20 = 60$
$x_1 + x_2 + x_3 + x_4 = 60 + 20$
$x_1 + x_2 + x_3 + x_4 = 80$
Thus,the sum of their present ages is $80$ years.

(B) The standard form of a linear equation in two variables is given by $ax + by + c = 0$.
Given the equation: $5y = -2x + 3$.
To convert this into the standard form,we move all terms to the left side of the equation.
Adding $2x$ to both sides: $2x + 5y = 3$.
Subtracting $3$ from both sides: $2x + 5y - 3 = 0$.
Comparing this with $ax + by + c = 0$,we get the standard form as $2x + 5y - 3 = 0$.

(A) two-digit number can be expressed as $10 \times (\text{digit at ten's place}) + (\text{digit at unit's place})$.
Given:
Digit at unit's place $= 2x - 1$
Digit at ten's place $= 2x + 1$
Substituting these values into the formula:
Number $= 10(2x + 1) + (2x - 1)$
$= 20x + 10 + 2x - 1$
$= 22x + 9$
Thus, the number is $22x + 9$.

(C) Given equations are:
$2x + 3y + 5 = 0$ --- $(1)$
$4x + 6y + 10 = 0$ --- $(2)$
Dividing equation $(2)$ by $2$,we get:
$2x + 3y + 5 = 0$
Since both equations are identical,they represent the same line.
For a pair of coincident lines,there are infinitely many solutions.
Therefore,the solution set is an infinite set.

(C) The equation $x=0$ represents the $y$-axis.
The equation $y=0$ represents the $x$-axis.
The intersection point of the $x$-axis and the $y$-axis is the origin.
The coordinates of the origin are $(0,0)$.
Therefore,the intersection point of the lines $x=0$ and $y=0$ is $(0,0)$.

(D) Let the present ages of the five friends be $a_1, a_2, a_3, a_4,$ and $a_5$.
According to the problem,$y$ years ago,the sum of their ages was $x$.
So,$(a_1 - y) + (a_2 - y) + (a_3 - y) + (a_4 - y) + (a_5 - y) = x$.
This simplifies to $(a_1 + a_2 + a_3 + a_4 + a_5) - 5y = x$.
The sum of their present ages is $(a_1 + a_2 + a_3 + a_4 + a_5) = x + 5y$.

(A) Let the present ages of Sachin and Sehwag be $S$ and $W$ respectively.
According to the problem,after $x$ years,the sum of their ages will be $y$ years.
So,$(S + x) + (W + x) = y$.
$S + W + 2x = y$.
Therefore,the present sum of their ages is $S + W = y - 2x$.
We need to find the sum of their ages $3$ years ago.
Sum of ages $3$ years ago $= (S - 3) + (W - 3) = (S + W) - 6$.
Substituting the value of $(S + W)$,we get $(y - 2x) - 6 = y - 2x - 6$ years.

(A) Let the cost of one table be $x$ and the cost of one chair be $y$.
According to the problem,we have two linear equations:
$3x + 2y = 3400$ --- (Equation $1$)
$2x + 3y = 3100$ --- (Equation $2$)
Adding Equation $1$ and Equation $2$:
$(3x + 2y) + (2x + 3y) = 3400 + 3100$
$5x + 5y = 6500$
Dividing the entire equation by $5$:
$x + y = 1300$
Thus,the total cost of one table and one chair is ₹ $1300$.

(A) $1$. Mahesh's age $x$ years ago was $y$ years.
$2$. Therefore,his current age is $(y + x)$ years.
$3$. After $z$ years,his age will be his current age plus $z$.
$4$. Age after $z$ years = $(y + x) + z = y + x + z$ years.

(B) two-digit number with $x$ at the ten's place and $y$ at the unit's place is represented as $10x + y$.
When a zero is inserted between these two digits,the new number becomes a three-digit number.
The digit $x$ moves to the hundred's place,the digit $0$ moves to the ten's place,and the digit $y$ remains at the unit's place.
Therefore,the value of the new number is $100(x) + 10(0) + y = 100x + y$.

(C) two-digit number with digit $y$ at the ten's place and digit $x$ at the unit's place is represented as $10y + x$.
According to the problem,this number is three times the sum of its digits.
The sum of the digits is $(x + y)$.
Therefore,the equation is $10y + x = 3(x + y)$.

(B) For a pair of linear equations $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$ to have infinite solutions,the condition is $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$.
Given equations are $3x + ky - 9 = 0$ and $x + 2y - 3 = 0$.
Here,$a_1 = 3, b_1 = k, c_1 = -9$ and $a_2 = 1, b_2 = 2, c_2 = -3$.
Applying the condition: $\frac{3}{1} = \frac{k}{2} = \frac{-9}{-3}$.
This simplifies to $3 = \frac{k}{2} = 3$.
From $\frac{k}{2} = 3$,we get $k = 3 \times 2 = 6$.

(C) The interest on an amount $x$ at $5 \%$ rate is given by $\frac{5}{100} \times x = 0.05x$.
Similarly,the interest on an amount $y$ at $7 \%$ rate is given by $\frac{7}{100} \times y = 0.07y$.
The total yearly interest is ₹ $310$.
Therefore,the equation is $0.05x + 0.07y = 310$.
Multiplying the entire equation by $100$ to remove the decimals,we get:
$5x + 7y = 31000$.

(A) Let the ages of Ina,Mina,and Dika be $I, M,$ and $D$ respectively.
The average age is given by $\frac{I + M + D}{3} = x$.
Therefore,the sum of their current ages is $I + M + D = 3x$.
After $y$ years,each person's age will increase by $y$ years.
The new sum of their ages will be $(I + y) + (M + y) + (D + y) = (I + M + D) + 3y$.
Substituting the sum of current ages: $3x + 3y$.
The new average age will be $\frac{3x + 3y}{3} = x + y$.

(A) The given equations are $x = -2$ and $y = 3$.
These equations represent two lines in the Cartesian plane.
The line $x = -2$ is a vertical line passing through the point $(-2, 0)$ on the $x$-axis.
The line $y = 3$ is a horizontal line passing through the point $(0, 3)$ on the $y$-axis.
The intersection point of these two lines is the unique solution to the system.
By substituting the values,the point of intersection is $(-2, 3)$.
Since a solution set is represented as a set containing the coordinate pair,the correct representation is $\{(-2, 3)\}$.

(C) To determine if $(-1, 3)$ is a solution to an equation,we substitute $x = -1$ and $y = 3$ into each equation:
$A) x - y + 4 = (-1) - (3) + 4 = -4 + 4 = 0$. This is a solution.
$B) 3x + y = 3(-1) + 3 = -3 + 3 = 0$. This is a solution.
$C) x + 3y + 8 = (-1) + 3(3) + 8 = -1 + 9 + 8 = 16 \neq 0$. This is not a solution.
$D) x + 2y - 5 = (-1) + 2(3) - 5 = -1 + 6 - 5 = 0$. This is a solution.
Therefore,the correct option is $C$.

(A) To determine if a point $(x, y)$ lies on the graph of the equation $2x - y = 1$,we substitute the $x$ and $y$ coordinates of the point into the equation.
If the left-hand side equals the right-hand side,the point lies on the graph.
$A) (0, 0): 2(0) - 0 = 0 \neq 1$. Thus,the graph does not pass through $(0, 0)$.
$B) (1, 1): 2(1) - 1 = 2 - 1 = 1$. The graph passes through $(1, 1)$.
$C) (3, 5): 2(3) - 5 = 6 - 5 = 1$. The graph passes through $(3, 5)$.
$D) (2, 3): 2(2) - 3 = 4 - 3 = 1$. The graph passes through $(2, 3)$.
Therefore,the graph does not pass through $(0, 0)$.

(B) Given the linear equation in two variables: $2x + 3y = 8$.
To express $y$ in terms of $x$,we isolate the term containing $y$ on one side of the equation.
Subtract $2x$ from both sides: $3y = 8 - 2x$.
Now,divide both sides by $3$ to solve for $y$: $y = \frac{8 - 2x}{3}$.
Thus,the correct expression is $\frac{8 - 2x}{3}$.

(C) Given the equation: $4x - 12y = 20$.
Divide the entire equation by $4$:
$\frac{4x}{4} - \frac{12y}{4} = \frac{20}{4}$
$x - 3y = 5$.
We need to find the value of $5x - 15y$.
Factor out $5$ from the expression:
$5x - 15y = 5(x - 3y)$.
Substitute the value of $(x - 3y) = 5$ into the expression:
$5(5) = 25$.
Therefore,$5x - 15y = 25$.

(C) line perpendicular to the $Y-axis$ is a horizontal line.
Any horizontal line has the general form $y = k$,where $k$ is a constant.
Let us analyze the given options:
$A) 3x + 2y = 5$ represents a slanted line.
$B) x = 3$ represents a vertical line (perpendicular to the $X-axis$).
$C) 2y - 3 = 0$ can be rewritten as $2y = 3$ or $y = 1.5$. This is a horizontal line,which is perpendicular to the $Y-axis$.
$D) x - y = 0$ represents a slanted line passing through the origin.
Therefore,the correct option is $C$.

(C) line perpendicular to the $X$-axis is a vertical line,which is represented by the equation of the form $x = k$,where $k$ is a constant.
In the given options:
$A) x - y = 0 \implies y = x$ (a line passing through the origin with slope $1$)
$B) x + y = 0 \implies y = -x$ (a line passing through the origin with slope $-1$)
$C) 3x + 2 = 0 \implies 3x = -2 \implies x = -2/3$ (this is a vertical line,perpendicular to the $X$-axis)
$D) 2y - 3 = 0 \implies 2y = 3 \implies y = 3/2$ (this is a horizontal line,parallel to the $X$-axis)
Therefore,the correct option is $C$.

(B) line is parallel to the $Y$-axis if its equation is of the form $x = k$,where $k$ is a constant.
Checking the options:
$A$) $3y - 1 = 0 \implies y = 1/3$ (This is a line parallel to the $X$-axis).
$B$) $2x + 5 = 0 \implies 2x = -5 \implies x = -2.5$ (This is a line parallel to the $Y$-axis).
$C$) $2x + 3y = 1$ (This is a slanted line intersecting both axes).
$D$) $2x - 3y = 1$ (This is a slanted line intersecting both axes).
Therefore,the correct option is $B$.

(D) line parallel to the $X$-axis is of the form $y = k$,where $k$ is a constant.
This is because for any value of $x$,the $y$-coordinate remains the same,resulting in a horizontal line.
Analyzing the given options:
$A$) $x=y$ represents a line passing through the origin with a slope of $1$.
$B$) $2x-y=0$ or $y=2x$ represents a line passing through the origin with a slope of $2$.
$C$) $x-2y=0$ or $y=x/2$ represents a line passing through the origin with a slope of $0.5$.
$D$) $2y=3$ can be written as $y = 1.5$,which is a constant value for $y$ regardless of $x$.
Therefore,the graph of $2y=3$ is a horizontal line parallel to the $X$-axis.

(A) linear equation in two variables $ax + by + c = 0$ passes through the origin $(0, 0)$ if and only if the constant term $c$ is equal to $0$.
In the given options:
$A) 4x = 5y \implies 4x - 5y = 0$. Here,$c = 0$,so it passes through the origin.
$B) 4x + 1 = 5y \implies 4x - 5y + 1 = 0$. Here,$c = 1 \neq 0$.
$C) 5x - 1 = 4y \implies 5x - 4y - 1 = 0$. Here,$c = -1 \neq 0$.
$D) 4x + 5y = 1 \implies 4x + 5y - 1 = 0$. Here,$c = -1 \neq 0$.
Therefore,the equation $4x = 5y$ passes through the origin.

(D) Let the two-digit number be represented as $10x + y$,where $x$ is the digit at the ten's place and $y$ is the digit at the unit's place.
Given that the digit at the ten's place is $x = 4$.
The product of the digits is $x \times y = 4y$.
According to the problem,the product of the digits is four times the digit at the ten's place:
$x \times y = 4 \times x$
Substitute $x = 4$ into the equation:
$4 \times y = 4 \times 4$
$4y = 16$
$y = 4$
Therefore,the digit at the unit's place is $4$.
The number is $10x + y = 10(4) + 4 = 44$.