Grade 9 Linear Equations in Two Variables In

Class 9Linear Equations in Two Variables

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Answer the questions

(1) A telecom operator charges Rs. 0.8 for the first minute and Rs. 0.5 per minute for subsequentminutes of a call. If duration of call is represented as x, and amount charged is represented as y,find the linear equation for this relationship.

(2) If solutions of a linear equation are (-9, -9), (0, 0) and (9, 9), find the equation.

Choose correct answer(s) from given choice

(3) A line passe through points (-2, 1) and (1, -2). Find the x-intercept of the line.a. -1 b. -2c. 0 d. -1.5

(4) The graph of equation for the line x = b is a linea. making an intercept b on both the axesb. parallel to x-axis at a distance b units from the originc. making an intercept b on the y-axisd. parallel to y-axis at a distance b units from the origin

(5) A point on line x = 0 is of the forma. (a, a) b. (0, a)c. (a, -a) d. (a, 0)

(6) The Autorickshaw fare in Pune is Rs. 18 for the first kilometer and Rs. 10 per kilometer forsubsequent distance covered. If distance is represented as d, and fare is represented as f, find thelinear equation for this relationship.a. f = 10d + 8 b. f = 18d + 8c. f = 18d + 10 d. f = 10d + 18

ID : in-9-Linear-Equations-in-Two-Variables [1]

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(7) Find the linear equation represented in the graph below.

a. y = -x b. y = xc. y = 0 d. y = -x - 1

(8) A line passe through points (2, -3) and (1, -2). Find the y-intercept of the line.a. -1 b. -3c. 0 d. -2

(9) The equation of x-axis is

a. x = y b. x = 0c. y = 0 d. x + y = 0

(10) The equation of straight line which is parallel to y-axis, and is at a distance of a from y-axis isa. y + x = a b. y = ac. x = a d. y - x = a

(11) In graph of linear equation 3x + 2y = 22, there is a point such that its ordinate is 4 less than itsabscissa. Find coordinates of the point.a. (5, 1) b. (6, 2)c. (2, 6) d. (1, 5)



(12) Equation 1x + 2y = 3 has a unique solution if x and y area. Real Numbers b. Positive Real Numbersc. Rational Numbers d. Natural Numbers

(13) A point of the form (a, -a) lies on the linea. x = 0 b. x = yc. y = 0 d. x + y = 0

(14) At what point does the line represented by the equation 3x + 8y = 31 intersects a line which isparallel to the y-axis, and at a distance 5 units from the origin and in the positive direction of x-axis.a. (2, 5) b. (5, 2)c. (3, 0) d. (0, 8)

Check True/False

(15) Equation 4x + 5y = 10 has a unique solution. True False

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Answers(1) y = 0.5x + 0.3

Step 1We are given the following facts- The charge for the first minute is is Rs. 0.8 - The charge per minute after that is Rs. 0.5Step 2We can see that the charge will be dependent on the time spent in minutesSo we set y on the left hand sidey = Some linear function of xStep 3We know that after the first minute, the rate is Rs. 0.5 per minute.So if the call lasts for x minutes, there will be a charge Rs. 0.8 for the first minute, and Rs. 0.5for x - 1 minutes.Step 4This means the equation is y = 0.8 + ((x - 1) x 0.5)Step 5Simplifying, we get y = 0.5x + 0.3

(2) x = yThe points (a,b) that solve a linear equation would satisfy ax+by=c, where a,b,c are constants. Substituting (0,0) we see that c = 0.Substituting the other two points (-9, -9) and (9, 9), we get the following equations-9x + (-9y)=0 and 9x + 9y = 0From these two equations, we can therefore eliminate the variables and see that the answer isx = y



(3) a. -1Step 1Equation of line y = m x + cStep 2Substitute first point in the equation1 = -2 m + cm = (1 - c)/-2 ________________(1)Step 3Substitute second point in the equation-2 = 1 m + cm = (-2 - c)/1 ________________(2)Step 4On equating value of m from both equations,(1 - c)/-2 = (-2 - c)/11 - 1c = 4 + 2c-3 c = 3c = -1Step 5m = (1 - c)/-2 = -1Step 6Equation of line : y = -1 x + cNow when line intersect with x axis, value of y will be 00 = -1x + (-1)x = -1

(4) d. parallel to y-axis at a distance b units from the originIf the equation for the line is x = b, this implies that the value of x is always b irrespective of thevalue of y.What this means is that this line is parallel to y-axis at a distance b units from the origin.



(5) b. (0, a)Try and trace the line x = 0 it the graph shown here

You can see that any point on the line defined by the equation x = 0 will always have the valueof x equal to 0. Therefore a point on the line will have the form of (0, a)

(6) a. f = 10d + 8Step 1We are given the following facts- The fare for the first kilometer is Rs. 18 - The fare per kilometer after that is Rs. 10Step 2We can see that the fare will be dependent on the distanceSo we set f on the left hand sidef = Some linear function of dStep 3We know that after the first kilometer, the rate is Rs. 10 per kilometer.So if we travel d kilometers, we will get charged Rs. 18 for the first kilometer, and Rs. 10 for d -1 kilometers.Step 4This means the equation is f = 18 + ((d - 1) x 10)Step 5Simplifying, we get f = 10d + 8



(7) a. y = -xThe general equation of a line is y=mx+cSo we have to find m and cTo find c, note from the equation that c is the value of y when x=0 (i.e. the equation becomesy=m*0 + c, or y=c).Look at the graph to see if this is a vertical line. If it is not (we'll see the case where it is later inthis tip), then what the value of y is when the equation crosses the vertical axisWe see that the value of y at this point is 0. So c=0The next part is finding mThe best way to consider m is to think of it as the slope of the line.Think of it as the change in y for a given change in x.Consider the two equations, y1 = mx1 + c, and y2 = mx2 + cNow we subtract the first equation from the secondWe get y1 - y2 = mx1 + c - (mx2 + c)Simplifying, (y1 - y2) = m(x1 - x2)or m = (y1 - y2)/(x1 - x2)Now, substitute the two points seen in the graph. m = (-2 - (1))/(2 - (-1))Also, note that this is the reason why we don't apply this when the line is vertical, because thedenominator would be 0, and the equation is meaninglessThis is solved to get the value of m, and get the answer m=-1

Now, if the line is a vertical one, then you can solve it by inspection.

So the answer is y= -x.



(8) a. -1Step 1Equation of line y = m x + cStep 2Substitute first point in the equation-3 = 2 m + cm = (-3 - c)/2 ________________(1)Step 3Substitute second point in the equation-2 = 1 m + cm = (-2 - c)/1 ________________(2)Step 4On equating value of m from both equations,(-3 - c)/2 = (-2 - c)/1-3 - 1c = -4 - 2c1 c = -1c = -1

(9) c. y = 0Take a look at a graph

You can see that for the x-axis, the value of y is always 0. So the equation is y=0



(10) c. x = aIf a line is parallel to the y-axis, then x value of it is constant for all values of y. Take a look at the image to see this case

Further, if the line is distance a away from the y-axis, it also means that this constant value of xis a. So the equation for that line is x=a

(11) b. (6, 2)Step 1We are given the following facts:The equation is 3x + 2y = 22The line has a point where the value of the ordinate is 4 less the value of the abscissaThe second fact implies the point is of the form (x,x - 4)Step 2Substituting this into the equation, we get3x + 2(x - 4) = 22Step 3Solving for this gets us the value of x = 6.From this we can find y = x - 4 = 2

(12) d. Natural NumbersA general equation in two variables has infinitely many solutions if there is no restriction placedon the values of the two variables (x and y here). However, it may have a unique solution ifcertain constraints are placed on it. Here we can see by observation that if x and y areconstrained to be natural numbers, then it has a solution for x=y=1, and this is the only possiblesolution for natural numbers.



(13) d. x + y = 0There are of course, infinite lines that can pass through a given point, but we have to choosefrom the four possibilities presented. The point specified is ((a, -a)). Out of the four options theonly one it actually can match is x + y = 0

(14) b. (5, 2)Let's consider the second line first.The line which is parallel to the y-axis and is at a distance 5 units from the origin in the positivedirection of the x-axis is defined by the following equation x=5 So, now we know that at the point of intersection, the value of x = 5The equation of the first line is 3x + 8y = 31Subtituting for x with the value 5 in this equation, we get y = 5So the answer is that the intersection is at the point (5, 2)

(15) FalseStep 1For two variables, we need at least two equations to find a unique solutionStep 2Therefore this one equation has infinitely many solutions



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Grade 9 Linear Equations in Two Variables In