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Simultaneous pair of Linear equation |
Condition |
Graphical representation |
Algebraic interpretation |
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$a_1x+b_1y+c_1=0$ $a_2x+b_2y+c_2=0$ Example $x-4y+14=0$ $3x+2y-14=0$ |
$\frac{a_{1}}{a_{2}}\neq \frac{b_{1}}{b_{2}}$ |
Intersecting lines. The intersecting point coordinate is the only solution
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One unique solution only. |
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$a_1x+b_1y+c_1=0$ $a_2x+b_2y+c_2=0$ Example $2x+4y=16$ $3x+6y=24$ |
$\frac{a_{1}}{a_{2}}= \frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$ |
Coincident lines. The any coordinate on the line is the solution.
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Infinite solution. |
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$a_1x+b_1y+c_1=0$ $a_2x+b_2y+c_2=0$ Example $2x+4y=6$ $4x+8y=18$ |
$\frac{a_{1}}{a_{2}}= \frac{b_{1}}{b_{2}}\neq \frac{c_{1}}{c_{2}}$ |
Parallel Lines
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No solution |
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S.no |
Type of method |
Working of method |
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1 |
Method of elimination by substitution |
(1) Suppose the equation are $a_1x+b_1y+c_1=0$ $a_2x+b_2y+c_2=0$ (2) Find the value of variable of either x or y in other variable term in first equation (3) Substitute the value of that variable in second equation (4) Now this is a linear equation in one variable. Find the value of the variable (5) Substitute this value in first equation and get the second variable |
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2 |
Method of elimination by equating the coefficients |
(1) Suppose the equation are $a_1x+b_1y+c_1=0$ $a_2x+b_2y+c_2=0$ (2) Find the LCM of a1 and a2 .Let it k. (3) Multiple the first equation by the value k/a1 (4) Multiple the first equation by the value k/a2 (5) Subtract the equation obtained. This way one variable will be eliminated and we can solve to get the value of variable y (6) Substitute this value in first equation and get the second variable |
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3 |
Cross Multiplication method |
(1) Suppose the equation are $a_1x+b_1y+c_1=0$ $a_2x+b_2y+c_2=0$ (2) This can be written as  (3) This can be written as  (4) Value of x and y can be find using the x => first and last expression y=> second and last expression |
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Condition |
Algebraic interpretation |
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$\frac{a_{1}}{a_{2}}\neq \frac{b_{1}}{b_{2}}$ |
One unique solution only. |
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$\frac{a_{1}}{a_{2}}= \frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$ |
Infinite solution. |
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$\frac{a_{1}}{a_{2}}= \frac{b_{1}}{b_{2}}\neq \frac{c_{1}}{c_{2}}$ |
No solution |
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Problem type |
Steps to be followed |
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Problem based on Distance, speed and time. |
We need to remember the Speed formula in these problem to derive the correct mathematical statement $\text {Speed} = \frac {\text{Distance}}{ \text{Time}}$ Or $\text{Distance} = \text {Speed} \times \text{Time}$ Or $\text{Time} = \frac {\text{Distance}}{ \text {Speed}}$ |
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Problem Based on moving upstream and downstream in the river |
We need to generally calculate the speed of Boat in still water and Speed of the stream. Let’s us assume them as x and y, then For downstream movement, speed of the boat would $= x+y$ For upstream movement, speed of the boat would $=x-y$ |
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Problem based on Geometry and Mensuration |
We need to remember the Angle sum property for Triangle, quadrilaterals, cyclic quadrilateral, parallelogram For Triangle, Sum of angles =1800 For quadrilateral, Sum of angles=3600 Similarly In case of parallelogram which is a special case of quadrilateral, the opposite angles are equal In case of cyclic quadrilaterals, the opposite angles are supplementary |
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Problem based on Numbers and digits. |
A two-digit number (xy y -in unit place and x in tenth place) can be expressed as 10x+y If the number are interchanged, it will be expressed as $10y+x$ So you can formulate the mathematical expression in the unknown using the above formula For three-digit number(xyz), the expression would $(100x+10y+z)$ |
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Problem based on Work Rate |
The standard equation that will be needed for these problems is Portion of Job done in given time= (Work rate of the person) X (Time taken working) |
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Age Problems |
It is good to take present age of the person as x and y. Then k year ago, there age would be (x-k) and (y-k) respectively Similarly, k year after, their age would be (x+k) and (y+k) respectively So you can formulate the mathematical expression based on this and then solve the problem |
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Fractions problem |
For Fraction problem, if the numerator is x and denominator is y, then the fraction is x/y. We can use the above expression to formulate the mathematical expressions |
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Mixing solution problems |
In this problem type, we will be looking at mixing solutions of different percentages to get a new percentage. The solution will consist of a secondary liquid (acid or alcohol) mixed in with water. Amount of secondary liquid will be given by Amount of secondary liquid= (percentage of secondary liquid) X (Volume of the solution) |
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Commercial mathematics problems |
Commercial mathematics problems include interest rate, cost price, selling price problem. Selling price or Marked price = Cost price + Profit So x be the selling price and y be the cost price Profit = y-x In case selling at loss, loss would be =x-y Similarly, simple interest problem could be formulated Simple interest = (% Rate of interest) X (number of year) X (Principle amount) |
Given below are the links of some of the reference books for class 10 math.
You can use above books for extra knowledge and practicing different questions.
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