We start with the equations $Y=X+1$ and $Y=X^2-3X-11$. To find the intersection points, set the two equations equal: $X^2-3X-11=X+1$. This simplifies to the quadratic equation $X^2-4X-12=0$.

To solve this quadratic equation, we use factorization. We need two numbers that multiply to $-12$ and add to $-4$. These numbers are $-6$ and $2$. Thus, we can factor the equation as $(X+2)(X-6)=0$.

Setting each factor to zero gives us the solutions for $X$: $X+2=0$ leads to $X=-2$, and $X-6=0$ leads to $X=6$. Next, we substitute these $X$ values back into the linear equation $Y=X+1$ to find the corresponding $Y$ coordinates.

For $X=-2$, $Y=-2+1=-1$. For $X=6$, $Y=6+1=7$. Therefore, the intersection points are $(-2,-1)$ and $(6,7)$.

• ★Set the equations equal to find intersections: $X^2-4X-12=0$0:03
• ★Factorization is preferred over the quadratic formula for solving quadratics.0:21
• ★The factors of the quadratic are $(X+2)(X-6)=0$.0:33
• ★Solutions for $X$ are $-2$ and $6$.0:45
• ★Substitute $X$ into $Y=X+1$ to find $Y$ coordinates.0:57
• ★Final intersection points are $(-2,-1)$ and $(6,7)$.1:10

Example: The intersection points of $Y=X+1$ and $Y=X^2-3X-11$ are $(-2,-1)$ and $(6,7)$.1:10

The expression $(x+p{)}^2$ expands to $x^2+2px+p^2$. Collecting like terms is necessary for simplification.

Equating coefficients involves setting the coefficients of corresponding terms equal. The coefficient of $x$ in the equation is 8, leading to $2p=8$, which gives $p=4$. The constant term gives the equation $10=p^2+q$. Substituting $p=4$ into $p^2$ gives $16$, leading to $10=16+q$, which simplifies to $q=-6$.

• ★The expression $(x+p{)}^2$ expands to $x^2+2px+p^2$.5:00
• ★Equating coefficients sets corresponding term coefficients equal.5:20
• ★From $2p=8$, we find $p=4$.5:40
• ★Substituting $p=4$ gives $q=-6$ from $10=16+q$.6:00

Example: Using $p=4$ and $q=-6$, we can write $x+4-6=30$.7:00

To solve the equation $x^2+4=36$, first add 6 to both sides, resulting in $x^2=36$. Next, take the square root of both sides, yielding $x+4=\text{±}6$. This gives two equations: $x+4=6$ and $x+4=-6$.

From $x+4=6$, subtract 4 to find $x=2$. From $x+4=-6$, subtract 4 to find $x=-10$. Thus, the two solutions are $x=2$ and $x=-10$.

Next, we translate the word problem about Gita buying fruits into an equation. Let $X$ be the number of apples, $X+7$ be the number of oranges, and $2X-1$ be the number of bananas. The total is 30, leading to the equation $X+(X+7)+(2X-1)=30$.

Combining like terms gives $4X+6=30$. Subtract 6 from both sides to get $4X=24$. Dividing by 4 results in $X=6$, meaning Gita buys 6 apples.

• ★The square root of 36 is both +6 and -6.7:27
• ★Translating word problems into equations simplifies solving.7:27
• ★The equation for total fruits is $X+(X+7)+(2X-1)=30$.7:27
• ★Combining terms leads to $4X+6=30$.7:27
• ★Solving gives $X=6$ apples.7:27

Example: Gita buys 6 apples, 13 oranges, and 11 bananas.7:27

Let $B$ represent the cost of one banana, which is given as $B=555$. To simplify the equation, calculate $15\times 6=90$ and $18\times 13=234$. The equation becomes $11B+324=555$. Subtracting 324 from both sides gives $11B=231$. Dividing both sides by 11 results in $B=21$.

Next, consider the equation $\frac{3w}{16}-1=\frac{1}{2}$. To eliminate the -1, add 1 to both sides, yielding $\frac{3w}{16}=\frac{3}{2}$. Cross-multiplying gives $3w\times 2=6w$ and $16\times 3=48$. Dividing both sides by 6 leads to $W=8$.

For the equation $2^{-y}=8$, recognize that $8$ can be expressed as $2^3$. This means $2^{-y}=2^3$. Since the bases are the same, set the exponents equal: $-y=3$. Therefore, $y=-3$.

• ★The cost of one banana is $B=21$12:26
• ★Cross-multiplication simplifies fractions effectively12:37
• ★$2^{-y}=8$ leads to $y=-3$13:14

Example: Solving $3w/16-1=1/2$ gives $W=8$.12:37

To solve simultaneous equations, we can use the elimination method. We add the equations together when one variable has opposite coefficients, such as $+Q$ and $-Q$. This results in a simplified equation: $2p+p=3p$.

Next, we simplify the constant terms. For example, $2-\frac{1}{2}$ can be rewritten as $\frac{4}{2}-\frac{1}{2}=\frac{3}{2}$. Multiplying both sides by 2 eliminates the fraction, giving us $3=6p$, leading to $p=\frac{1}{2}$ or $0.5$. We then substitute $p$ back into one of the original equations to find $Q$.

• ★Use elimination to simplify simultaneous equations.17:25
• ★Adding equations with opposite coefficients simplifies the problem.17:30
• ★Keep results in fraction form for clarity.17:35
• ★Substituting back helps find the second variable.17:45

Example: Substituting $p=\frac{1}{2}$ into $p-Q=-\frac{1}{2}$ gives $Q=1$.17:55

To find angles using inverse trigonometric functions, we start with the sine function. The equation $\sin u=\frac{1}{2}$ leads us to use the inverse sine function, giving us $u={\sin }^{-1}\left(\frac{1}{2}\right)$. This results in $u=30$ degrees, but we must recognize that there are two possible angles.

The sine graph peaks at 90 degrees and has a periodic nature. The second angle is found by calculating $180-30$, which gives us $150$ degrees. Similarly, for the cosine function, $\cos v=1$ leads to $v={\cos }^{-1}(1)$, resulting in $v=0$ degrees, but this is not the only solution. The cosine graph starts at 1 and decreases to -1 at 180 degrees.

• ★The inverse sine function gives angles for a sine value.20:25
• ★For ${\sin }^{-1}\left(\frac{1}{2}\right)$, the angles are $30$ and $150$ degrees.20:35
• ★The inverse cosine function gives angles for a cosine value.20:45
• ★${\cos }^{-1}(1)$ results in $0$ degrees, but more solutions exist.20:55

Example: For $\sin u=\frac{1}{2}$, we find $u=30$ degrees and $150$ degrees.20:30

To simplify expressions, collect like terms. For example, $1A-3A=-2A$ and $-2B+7B=5B$. Be cautious with negative numbers.

Use the distributive property correctly. For instance, $4$ times $x$ equals $4X$, and $4$ times $-5$ equals $-20$. Remember that multiplying two negatives results in a positive: $-1$ times $-2x=+2x$.

When working with algebraic fractions, create a common denominator by multiplying the denominators together. For example, the common denominator for $2X$ and $x-5$ is $2X(x-5)$.

To solve equations, perform inverse operations to isolate the variable. For example, from $13-4x=6-x$, add $4x$ to both sides to simplify.

• ★Collect like terms to simplify expressions, e.g., $1A-3A=-2A$22:52
• ★Be cautious with negative numbers when simplifying, e.g., $-2B+7B=5B$22:55
• ★Use the distributive property correctly, e.g., $4$ times $-5=-20$22:58
• ★Recognize that multiplying two negatives results in a positive, e.g., $-1$ times $-2x=+2x$23:01
• ★To create a common denominator for fractions, multiply the denominators together23:05
• ★When solving equations, perform inverse operations to isolate the variable23:10

Example: Solve the equation $13-4x=6-x$ to find $x=-5$.23:25

To make $X$ the subject of the formula, we start by eliminating fractions. Multiply both sides by $X$ to get rid of the division. This gives us $XY=5(P-2X)$.

Next, expand the brackets: $5(P-2X)$ becomes $5P-10X$. Rearrange the equation to have all terms with $X$ on one side: $XY+10X=5P$. Factor out $X$ to get $X(Y+10)=5P$. Finally, isolate $X$ by dividing by $(Y+10)$, leading to $X=\frac{5P}{Y+10}$.

• ★Make $X$ the subject by isolating it.27:50
• ★Eliminate fractions by multiplying by $X$.27:50
• ★Expand brackets: $5(P-2X)=5P-10X$.27:50
• ★Isolate $X$ by dividing by $(Y+10)$.27:50

Example: From $XY=5(P-2X)$, we find $X=\frac{5P}{Y+10}$.27:50

Factorizing expressions can yield easy marks in exams, especially for higher grades. The technique of 'splitting the term' is useful for factorizing four-term expressions. For example, consider the expression $1-q-a+aq$. We can separate this into two parts: $1-q$ and $-a+aq$.

Next, we identify common factors. From $-a+aq$, we can factor out $a$, giving us $a(-1+q)$. Changing the signs allows us to rewrite this as $a(q-1)$. Now we have $1-q$ and $a(q-1)$, which share a common factor of $1-q$.

The final factorized form is $(1-q)(1-a)$. To verify, expand this back to the original expression. If you do this correctly, you will return to $1-q-a+aq$.

Let's look at another example: $2m+3p-8km-12kp$. We can split this into $2m+3p$ and $-8km-12kp$. The common factor here is $-4k$, which we factor out, leading to $-4k(2m+3p)$. The final factorized form is $(2m+3p)(1-4k)$.

• ★Factorizing can yield easy marks in exams.29:53
• ★Splitting the term helps in factorizing four-term expressions.29:55
• ★Identify common factors in the expression.30:05
• ★Changing signs can help achieve desired bracket forms.30:15
• ★Verify factorization by expanding back to the original expression.30:25
• ★Common factors can include numerical coefficients and variable terms.30:35
• ★Factorization can involve multiple steps, including grouping terms.30:45
• ★The concept of 'dots' introduces a different style of factorization.30:55

Example: For the expression $1-q-a+aq$, the factorized form is $(1-q)(1-a)$.30:20

The difference of two squares (d-o-t-s) allows us to factor expressions with square numbers into two brackets. For example, $x^2-4y^2$ can be factored as $(x-2y)(x+2y)$ by taking the square roots of each term.

To apply d-o-t-s, take the square root of each term and use one plus and one minus to eliminate cross terms. This technique is crucial for simplifying expressions efficiently.

When factorizing expressions with common factors, identify the greatest common factor. For instance, in $18px-27p$, the common factor is $9p$, leading to $9p(2x-3)$.

For expressions with four terms, rearranging can help find common factors. For example, rearranging $mt-m+nt-n$ allows us to factor it into $(t-1)(m+n)$.

• ★The difference of two squares factors as $(a-b)(a+b)$34:52
• ★Common factors can simplify expressions significantly.35:05
• ★Rearranging terms can reveal hidden common factors.35:15
• ★Factorizing algebraic fractions involves simplifying both numerator and denominator.35:25

Example: Factor $5(x^2-4)$ as $5(x-2)(x+2)$ using the difference of squares.35:10

Watch the video on trigonometric equations. It will clarify the upcoming question.

Understanding trigonometric equations will make the question easier to solve.

• ★Watch the trigonometric equations video.39:51
• ★Clarifying trigonometric equations aids in solving related questions.39:52
• ★The upcoming question will be easier after the trig video.39:53