SAT MATH: HEART OF ALGEBRA
Algebra, the art of knowing the unknown.
so point 1 (x,y) = (0,0)
let x = 1 => y = x = 1
thus the other point is (1,1)
now draw a line from these two points and that is the graph of
Y = X
SAT Math covers Heart of Algebra in it's syllabus, which is one of the easiest thing to understand, if you want to.
To understand Algebra, you need to master graphs thus we will go hand in hand with equations and their graphs.
LINE: Y = MX + C
PARABOLA: ax2 + bx = c
SAT Math focuses on the graphs of lines which you are probably gonna encounter a lot. So lets understand graphs,
X = 3
Y = 3
Y = X
Y = -X
If you still couldn't figure out how graphs are made then let me help you, Since the power of x and y is 1 thus the graph will be a straight line. A straight line is the shortest path joining 2 distinct points thus, take any 2 values of x and then find y and then plot them them in a graph and then draw a straight line joining them.
EX1) Y = X
if x = 0 => y = x = 0 => y = 0
so point 1 (x,y) = (0,0)
let x = 1 => y = x = 1
thus the other point is (1,1)
now draw a line from these two points and that is the graph of
Y = X
EX2) X = 3
Now in the graph of X = 3, the value of X is fixed but you can put any value of Y. Thus your points can be (3,0), (3,1).
so plot them and join them.
EX3) Y = 3X + 3
if X = 0 => Y = 3; if x = 1 => Y = 6
thus points are(0,3) and (1,6). Now plot them and join them.
Now let's work on creating linear equations from statements,
Q) A man creates 20 pots in 1 hour. If he has 500 pots and he has to reach a target of making 800 pots, how many hours will he have to work?
A) Let number of hours he works = X
and number of pots = Y
he already has 500 pots
thus Y = 500 + new pots
new pots = number of pots created in 1 hour ✕ number of hours
= 20 ✕ X
=> Y = 500 + 20X
target is of 800 pots thus number of pots = Y = 800
800 = 500 + 20X
800 - 500 = 20X
300 = 20X
15 = x
=> X = 15
so it will take him 15 hours to complete the target of 800 pots.
#EQUATION IN 2 VARIABLES
SO IN THIS SECTION WE WILL FIND THE SOLUTION OF 2 LINEAR EQUATION THROUGH VARIOUS METHODS.
X + Y = 3 --------(1)
3X + 2Y = 7 --------(2)
1)SUBSTITUTION
In this method we substitute the value of one variable derived from one equation into another one.
X + Y = 3
X = 3 - Y
Substituting the value of X in equation 2
3(3 - Y) + 2Y = 7
9 - Y = 7 => 2 = Y
X = 3 - Y = 3 - 2 = 1
So, X = 1, Y = 2
2) ELIMINATION
In Elimination method we subtract 1 equation from another in such a way that it gets free from a certain variable and it becomes an equation with 1 variable.
Equation 1 is
X + Y= 3
Multiplying by 3
3X + 3Y = 9
Subtracting equation 2 from this equation
3X + 3Y = 9
-(3X + 2Y = 7)
---------------------
0 + Y = 2
Substitute it in equation 1
X + 2 = 3 => X = 1
3)GRAPHICAL METHOD
You can draw these equations and see where they meet.
SOLVING EQUATIONS WITH X2
METHOD 1) SPLITTING THE MIDDLE TERM
Break the middle term into 2 terms which have same product that the product of first and last term have.
like in equation
So X = -2,-1
# NUMBER OF SOLUTIONS OF TWO EQUATIONS
1) If they meet at 1 point then 1 solution.
If you solve it and you get an answer then it has one solution
2) If they are one over another, then it has infinite number of solutions.
if you get a statement free from X and Y and is true such as 0 = 0 or 2 = 2 then it has infinitely many solutions
3) If they are parallel then no solution
if you get an absurd statement such as 1 = 3 or 2 = 4 then it has no solution.
#EQUATION IN 2 VARIABLES
SO IN THIS SECTION WE WILL FIND THE SOLUTION OF 2 LINEAR EQUATION THROUGH VARIOUS METHODS.
X + Y = 3 --------(1)
3X + 2Y = 7 --------(2)
1)SUBSTITUTION
In this method we substitute the value of one variable derived from one equation into another one.
X + Y = 3
X = 3 - Y
Substituting the value of X in equation 2
3(3 - Y) + 2Y = 7
9 - Y = 7 => 2 = Y
X = 3 - Y = 3 - 2 = 1
So, X = 1, Y = 2
2) ELIMINATION
In Elimination method we subtract 1 equation from another in such a way that it gets free from a certain variable and it becomes an equation with 1 variable.
Equation 1 is
X + Y= 3
Multiplying by 3
3X + 3Y = 9
Subtracting equation 2 from this equation
3X + 3Y = 9
-(3X + 2Y = 7)
---------------------
0 + Y = 2
Substitute it in equation 1
X + 2 = 3 => X = 1
3)GRAPHICAL METHOD
You can draw these equations and see where they meet.
SOLVING EQUATIONS WITH X2
METHOD 1) SPLITTING THE MIDDLE TERM
Break the middle term into 2 terms which have same product that the product of first and last term have.
like in equation
X2 + 3x + 2 = 0
X2 + 2X + X + 2 = 0
X(X + 2) + (X + 2) = 0
(X +1 )(X=2) = 0
=> X = -1, -2
METHOD 2) QUADRATIC FORMULA
CONSIDER THE EQUATION X2 + 3x + 2 = 0 IN THE FORM OF AX2 + Bx + C = 0
thus for this equation A= 1, B = 3, C = 2
FORMULA
So X = -2,-1
# NUMBER OF SOLUTIONS OF TWO EQUATIONS
1) If they meet at 1 point then 1 solution.
If you solve it and you get an answer then it has one solution
2) If they are one over another, then it has infinite number of solutions.
if you get a statement free from X and Y and is true such as 0 = 0 or 2 = 2 then it has infinitely many solutions
3) If they are parallel then no solution
if you get an absurd statement such as 1 = 3 or 2 = 4 then it has no solution.
INEQUALITY
While dealing with algebra, it's not necessary that you will always get equations with equality signs sometimes we get inequality ones denoting the greater and lesser side..
Let's see what a general inequality in one variable looks like.
AX + B > C
AX + B < C
EX) 3X + 1 > 7
ANS) Let's solve it
subtracting 1 from both sides.
3X > 6
Dividing by 3 on both sides
X > 2
So basically what we have to do is to take X on 1 side with no constant and make it's coefficient 1.
CREATING INEQUALITIES
EX) A man wants to buy a folder and as many packets of pages as many he can. If the cost of a folder is $8 and the cost of a packet of pages is $3. If the man has $20 then how many packets of pages can he buy.
ANS) The total amount of money he has = 20
so total money he spent must be less than equal to 20
Now, let the number of packets of pages be X
So, total cost = cost of a folder = X ✕ cost of a pen
Total cost ≤ 20
cost of a folder + X ✕ cost of pen ≤ 20
8 + 3X ≤ 20
3X ≤ 12
X ≤ 4
So max number of packets he can buy = 4
LINEAR INEQUALITY IN 2 VARIABLES
EX)You earn part time money at a bookstore and as a lifeguard. In the bookstore you are paid $10 per hour and as a lifeguard you are paid $25 per hour. You don't want want to work part time for more than 20 hrs a week as that will hinder your personal life and you want to work at both the places for a significant amount of time. You want to earn about $180 a week. Now since this is your life, so form suitable equations for them.
ANS) Let the number of hours you work in the bookstore = B
and the number of hours you work as a lifeguard = L
L + B ≤ 20
Total wealth to earn → ≥ $180
Wealth from bookstore + Wealth earned as Life Guard ≥ $180
$10 ✕ B + $25 ✕ L ≥ $180
So the equations are
L + B ≤ 20
2B + 5L ≥ 36 ( Inequality and equation are generally present in simplest form in SAT but exceptions are always there.)
SOLVING INEQUALITIES WITH 2 VARIABLES
EX) Which of the following ordered pair (X,Y) satisfies the system of inequalities below.
X - Y ≥ 2
X + 2Y > 4
A) (2,0)
B) (2,3)
C) (1,2)
D) (3,1)
ANS) In these questions you have to plug in the values of options in inequalities to find the answer.
Plugging option A
X - Y ≥ 2
2 - 0 ≥ 2
2 - 3 ≥ 2
-1 ≥ 2 => False
1 - 2 ≥ 2
-1 ≥ 2 => False
3 - 1 ≥ 2
2 ≥ 2 => True
Now in inequality 2
X + 2Y > 4
3 + 2 > 4
5 > 4 => TRUE
It may seem a bit lengthy but believe me with a bit of practice you will get used to it.
Total cost ≤ 20
cost of a folder + X ✕ cost of pen ≤ 20
8 + 3X ≤ 20
3X ≤ 12
X ≤ 4
So max number of packets he can buy = 4
LINEAR INEQUALITY IN 2 VARIABLES
EX)You earn part time money at a bookstore and as a lifeguard. In the bookstore you are paid $10 per hour and as a lifeguard you are paid $25 per hour. You don't want want to work part time for more than 20 hrs a week as that will hinder your personal life and you want to work at both the places for a significant amount of time. You want to earn about $180 a week. Now since this is your life, so form suitable equations for them.
ANS) Let the number of hours you work in the bookstore = B
and the number of hours you work as a lifeguard = L
L + B ≤ 20
Total wealth to earn → ≥ $180
Wealth from bookstore + Wealth earned as Life Guard ≥ $180
$10 ✕ B + $25 ✕ L ≥ $180
So the equations are
L + B ≤ 20
2B + 5L ≥ 36 ( Inequality and equation are generally present in simplest form in SAT but exceptions are always there.)
SOLVING INEQUALITIES WITH 2 VARIABLES
EX) Which of the following ordered pair (X,Y) satisfies the system of inequalities below.
X - Y ≥ 2
X + 2Y > 4
A) (2,0)
B) (2,3)
C) (1,2)
D) (3,1)
ANS) In these questions you have to plug in the values of options in inequalities to find the answer.
Plugging option A
X - Y ≥ 2
2 - 0 ≥ 2
2 ≥ 2 => TRUE
Now in inequality 2
X + 2Y > 4
2 + 0 > 4
2 > 4 => False
Plugging option B
X - Y ≥ 22 - 3 ≥ 2
-1 ≥ 2 => False
Plugging option C
X - Y ≥ 21 - 2 ≥ 2
-1 ≥ 2 => False
Plugging option d
X - Y ≥ 23 - 1 ≥ 2
2 ≥ 2 => True
Now in inequality 2
X + 2Y > 4
3 + 2 > 4
5 > 4 => TRUE
It may seem a bit lengthy but believe me with a bit of practice you will get used to it.
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