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All of Algebra in just over 1 Hour!! Foundation & Higher Grade 4-9 Maths Revision | GCSE Maths Tutor

The GCSE Maths Tutor・2 minutes read

Index laws involve adding and subtracting powers with the same base, dividing numbers first in division problems, and expanding and simplifying brackets to factorize expressions while solving equations by isolating variables using inverse operations. Quadratic equations are solved by completing the square, finding turning point coordinates, and exploring simultaneous equations, quadratic inequalities, and algebraic fractions. The text also covers graph plotting, finding coordinates, using the quadratic formula, and simplifying equations, including demonstrating the process of finding inverse functions and effect variations on coordinates.

Insights

Index laws dictate adding powers with the same base when multiplied and subtracting when divided, guiding calculations involving powers effectively.
Simplifying algebraic expressions involves expanding brackets, combining like terms, and factorizing to achieve a concise form, enhancing clarity and ease of computation.
Solving equations necessitates isolating variables through inverse operations, such as adding or subtracting common numbers and reversing multiplication or division, streamlining the process.
Graphing linear equations involves identifying gradients and y-intercepts to plot lines accurately, while parallel lines share the same gradient, simplifying graph interpretation.

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Recent questions

How are powers handled in division problems?
When dividing powers, numbers are divided first, then powers subtracted.

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Summary

00:00

Mathematical Operations and Simplification Techniques

Index laws involve adding powers with the same base when multiplied and subtracting when divided.
In a division problem with powers, numbers are divided first, then the powers are subtracted.
When dealing with powers in brackets, multiply the powers together.
Dividing numbers in a fraction involves dividing the numbers first and then subtracting the powers of the variables.
Expanding brackets involves multiplying each term inside the bracket by the term outside.
Simplifying brackets requires combining like terms and simplifying the expression.
Expanding and simplifying double brackets involves multiplying each term in one bracket by each term in the other bracket.
Factorizing involves taking out common factors and putting the expression into brackets.
Fully factorizing includes taking out all possible factors, not just one.
Solving equations requires isolating the variable by performing inverse operations.

11:54

Solving Equations and Word Problems with Algebra

To solve an equation, start by removing a common number from both sides, such as subtracting 5 from both sides to get Y - 5 = X/3.
To reverse a division by 3, multiply both sides by 3, resulting in 3(Y - 5) = X.
Expanding the equation, 3Y - 15 = X, or 3Y = X + 15.
When dealing with a fraction, multiply by the denominator to unlock the top part, like multiplying by 3 to get 3Y = X + 5.
Substituting values into equations, such as A = 5B + 2C, B = 3, and Y = -2, find the value of A by substituting and calculating.
When solving simultaneous equations, make the coefficients of a variable the same, like multiplying one equation by 3 and the other by 4 to eliminate variables.
Add or subtract the equations to eliminate variables and find the values, like adding equations to get 13X = 91, then solving for X to get X = 7.
Sub X = 7 back into an equation to find Y, like substituting into the first equation to find Y = -3.
Create equations based on word problems, like setting up equations for calculators and books costing a total amount, then solving to find the individual prices.
Use algebra to solve worded problems, like determining the ages of Adam, Brian, and Chris based on given relationships and the sum of their ages.

24:57

Mathematical Problem Solving Techniques Explained

The value of x is 5.5, considering the units specified in the question.
The perimeter of the triangle is calculated by substituting x into the sides, resulting in a total of 24 units.
The quadrilateral's perimeter is double that of the triangle, making it 40 units.
To find the angle ABC in the Pentagon, the total sum of angles in a Pentagon (540 degrees) is used.
The equation for the angles in the Pentagon is solved, leading to x equaling 73 degrees.
Angle ABC is then calculated as 2 times x, resulting in 146 degrees.
Solving an inequality is approached similarly to solving an equation, with x being less than 8.5.
The nth term for a sequence is determined by recognizing the pattern and adjusting it accordingly.
The 50th term in the sequence is found by applying the formula and calculating it to be 199.
Graphs of linear and quadratic equations are plotted by either creating a table of values or understanding the equation's components like gradient and intercept.

37:34

Graphing Lines and Equations: A Comprehensive Guide

Cubic, reciprocal, and other graph shapes can be plotted similarly by substituting numbers carefully within brackets.
For a straight line graph, find the equation in the form y = mx + c by identifying two coordinates and calculating the gradient (m) and y-intercept (c).
Example: y = 2x - 3 represents a line with a gradient of 2 and a y-intercept at -3.
To find the equation of a line with a different slope, repeat the process by determining the y-intercept and gradient.
Example: y = -2x + 1 signifies a line with a gradient of -2 and a y-intercept at 1.
Two parallel lines have the same gradient, identified by the equation in the form y = mx + c.
To determine parallel lines, ensure the gradients in the equations match.
Example: Lines with equations y = 3x and y = 3x + 5 are parallel due to both having a gradient of 3.
Expanding triple brackets involves multiplying each term carefully to simplify the expression.
Factorizing a quadratic with a coefficient greater than one requires finding factors that multiply to the constant term and add up to the coefficient of the x term.
Rearranging equations to isolate a variable involves steps like squaring both sides, eliminating denominators, and factorizing to achieve the desired variable as the subject.

49:54

Analyzing Quadratic Sequences and Graphs

The text discusses quadratic sequences related to square numbers, focusing on finding the N term of a sequence.
It emphasizes the difference between the N squared sequence and the given sequence, highlighting the need to identify the pattern.
The text explains the process of determining the N term of a quadratic sequence by analyzing the differences between numbers.
It provides a formula for finding the N term of a sequence when the differences are not constant.
The text then shifts to finding the coordinates of the turning point of a graph by completing the square.
It details the steps involved in completing the square to find the turning point coordinates.
The text further delves into solving simultaneous equations representing a circle and a line, leading to two sets of coordinates.
It explains the process of substituting one equation into the other to solve the simultaneous equations.
The text then explores quadratic inequalities, illustrating how to find the regions that satisfy specific inequalities on a graph.
It demonstrates how to shade the regions that satisfy multiple inequalities on a graph, considering the different scenarios presented.

01:02:25

Graphing, Inequalities, and Quadratic Equations Explained

The first line equation is 2x + 1, which is drawn on the graph.
The second line equation is x + y = 4, represented by a diagonal line connecting 4 on both axes.
The third line equation is y = 1, a straight line passing through y = 1.
To shade the region satisfying the inequalities, coordinates are tested in the equations.
Testing a coordinate like (1,2) in the inequalities helps determine the shaded region.
The quadratic equation is solved using the quadratic formula: (-B ± √(B² - 4AC)) / 2A.
Given a quadratic with a = 5, b = -11, and c = -13, the solutions are calculated to three significant figures.
Algebraic fractions are simplified by factorizing, canceling out common factors.
Adding algebraic fractions with different denominators requires finding a common denominator.
Solving equations involving algebraic fractions involves expanding and simplifying the expressions.

01:14:15

Solving Quadratic Equations and Inverse Functions

The equation given is 26X - 4 = 6X^2 + 2X - 4.
Rearranging the equation to equal 0 on the left side by subtracting 26X and adding 4 to both sides.
Simplifying the quadratic equation to 6X^2 - 26X - 24X + 4 = 0.
Factoring the equation to get the solutions: X = 0 and X = 4.
Explaining the process of solving quadratic equations and the importance of avoiding negative coefficients.
Simplifying the equation by dividing both sides by 6 to get 0 = X^2 - 4X.
Factorizing the simplified equation to get the solutions: X = 0 and X = 4.
Introducing functions f(X) = 2X - 4 and g(X) = X^2 + 5.
Demonstrating the process of finding the inverse function of f(X) as X = (Y + 4) / 2.
Explaining how to substitute one function into another to find the value of the composite function.

01:26:25

Coordinate Changes and Algebraic Patterns Explained

The opposite of X's effect is inside the bracket next to X, halving the x-coordinate instead of doubling it.
The y-coordinate remains unaffected by the X's opposite effect.
Different variations may require drawing graphs, but focusing on the coordinate changes can simplify the process.
When outside the bracket next to Y, the effect is as expected, affecting the y-coordinate.
Algebraic proof involves showing that the sum of the squares of two consecutive odd numbers is always two more than a multiple of eight.
Even numbers are represented as 2n, while odd numbers are 2n+1.
Squaring and summing two consecutive odd numbers, like 2n+1 and 2n+3, leads to a specific pattern.
Factorizing the sum of squares by 8 reveals that it is always two more than a multiple of eight.
Finding the equation of a perpendicular line passing through a specific point involves determining the gradient and solving for the y-intercept.

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