|
Simultaneous pair of Linear equation |
Condition |
Graphical representation |
Algebraic interpretation |
|
$a_1x+b_1y+c_1=0$ $a_2x+b_2y+c_2=0$ 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
|
One unique solution only. |
|
$a_1x+b_1y+c_1=0$ $a_2x+b_2y+c_2=0$ 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.
|
Infinite solution. |
|
$a_1x+b_1y+c_1=0$ $a_2x+b_2y+c_2=0$ 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
|
No solution |
|
S.no |
Type of method |
Working of method |
|
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 |
|
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 |
|
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 |
|
Condition |
Algebraic interpretation |
|
$\frac{a_{1}}{a_{2}}\neq \frac{b_{1}}{b_{2}}$ |
One unique solution only. |
|
$\frac{a_{1}}{a_{2}}= \frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$ |
Infinite solution. |
|
$\frac{a_{1}}{a_{2}}= \frac{b_{1}}{b_{2}}\neq \frac{c_{1}}{c_{2}}$ |
No solution |
|
Problem type |
Steps to be followed |
|
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 $\text{Time} = \frac {\text{Distance}}{ \text {Speed}}$ |
|
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 $=x-y$ |
|
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 |
|
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 $(100x+10y+z)$ |
|
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) |
|
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 |
|
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 |
|
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) |
|
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.
Thanks for visiting our website.
DISCLOSURE: THIS PAGE MAY CONTAIN AFFILIATE LINKS, MEANING I GET A COMMISSION IF YOU DECIDE TO MAKE A PURCHASE THROUGH MY LINKS, AT NO COST TO YOU. PLEASE READ MY DISCLOSURE FOR MORE INFO.