Algebraic Fractions (Equations) - GCSE Higher Maths

Summary

This instructional video teaches viewers how to solve equations involving algebraic fractions. It begins by combining fractions on the left side of an equation using a common denominator, then simplifies and solves for the variable. The video progresses through multiple examples of increasing difficulty, including those with quadratic equations that require factorization or the use of the quadratic formula. It concludes with solving equations with complex denominators and provides a step-by-step approach to finding the solutions. The video aims to equip learners with the skills to tackle a variety of algebraic fraction problems.

Takeaways

• 📚 The video is a tutorial on solving equations with algebraic fractions.
• 🔗 It builds upon a previous video titled 'algebraic fractions operations', which is linked in the description for reference.
• 📉 The initial step involves focusing on the left-hand side of the equation and combining fractions by finding a common denominator.
• 📌 To combine fractions, the video demonstrates converting each fraction to have a common denominator, often the least common multiple (LCM) of the original denominators.
• 📝 The process includes multiplying both the numerator and the denominator by appropriate factors to achieve the common denominator.
• ➗ After combining the fractions, the video shows how to expand and simplify the resulting expressions.
• 🔢 The video provides step-by-step solutions to various equations, including multiplying both sides by the denominator to eliminate fractions.
• 📈 It explains how to handle equations with quadratic expressions, including factorization and the use of the quadratic formula.
• 🎯 The video also covers how to deal with equations that result in a quadratic that does not factorize easily, requiring the quadratic formula or completing the square.
• 📊 For complex equations, the script illustrates the process of expanding brackets and simplifying terms before solving.
• 📑 The video concludes with a series of examples that incrementally increase in difficulty, guiding viewers through the problem-solving process.

Q & A

What is the main topic of the video?

The main topic of the video is solving equations that contain algebraic fractions.

What should viewers do if they are not familiar with algebraic fractions operations?

Viewers should check out the previous video titled 'algebraic fractions operations' which is linked in the description of the current video.

What is the first step in solving an equation with algebraic fractions according to the video?

The first step is to focus on the left-hand side of the equation and combine the fractions into one fraction over a common denominator.

How is the common denominator found for fractions with denominators 6 and 4?

The common denominator is found by determining the least common multiple of the denominators, which in this case is 12.

What is the process of rewriting fractions over a common denominator?

Each fraction is rewritten by multiplying both the numerator and the denominator by the factor needed to reach the common denominator.

How does the video handle the right-hand side of the equation initially?

The right-hand side of the equation is left alone initially while the left-hand side is combined into one fraction.

What mathematical operation is used to simplify the numerator after combining fractions?

The video uses expansion of brackets and then collects like terms to simplify the numerator.

Why is it necessary to multiply both sides of the equation by 12 after combining fractions?

Multiplying both sides by 12 eliminates the fraction and makes the equation simpler to solve.

What is the solution to the first example equation provided in the video?

The solution to the first example equation is x = 27/5, which is 5.4.

How does the video approach a more complex equation with denominators 6 and 9?

The video uses the same approach of finding a common denominator, which is the least common multiple of 6 and 9, and then combines the fractions on the left-hand side.

What is the common mistake to avoid when expanding brackets in equations with algebraic fractions?

The common mistake to avoid is incorrectly handling the signs when multiplying terms, especially with subtraction involved.

How does the video handle equations with quadratic expressions?

The video shows how to set the quadratic expression equal to zero and then solve for x using methods such as factorization or applying the quadratic formula.

What is the final form of the solution that the video demonstrates for a particularly tricky equation?

The final form of the solution for the tricky equation is x = a + b√13/b, where a and b are specific numerical values derived from the quadratic formula.

How does the video simplify the square root term in the quadratic formula?

The video simplifies the square root term by factoring out common factors and using the properties of square roots to express the term in its simplest form.

What is the process for solving the last equation in the video?

The process involves combining fractions over a common denominator, expanding and simplifying the numerator, multiplying both sides by the denominator to clear the fraction, and then solving the resulting quadratic equation.

What are the two solutions to the final equation presented in the video?

The two solutions to the final equation are x = 8.5 and x = 1/3.

Outlines

00:00

📚 Introduction to Solving Algebraic Fraction Equations

This paragraph introduces the topic of solving equations involving algebraic fractions. It suggests reviewing a previous video on algebraic fraction operations and outlines the process of combining fractions on the left-hand side of an equation by finding a common denominator. The example given demonstrates converting fractions with denominators 6 and 4 into a common denominator of 12, combining them, and then solving the resulting equation to find x = 5.4.

05:00

🔍 Expanding and Simplifying Equations with Algebraic Fractions

This section builds upon the initial approach by tackling a more complex equation with fractions having denominators of 6 and 9. The common denominator is determined to be 18, and the fractions are rewritten and combined accordingly. After expanding and simplifying the equation, the solution process involves multiplying both sides by 18 and isolating x to find x = 4.

10:04

📘 Solving Equations with Multi-term Denominators

The paragraph delves into solving equations where denominators have multiple terms, such as x + 2 and x + 5. The least common multiple of these is used as the common denominator, and the fractions are rewritten and combined. After expanding and simplifying, the equation is transformed into a quadratic equation. The process of solving this quadratic equation involves setting it to zero and factoring it to find two solutions, x = -4 and x = 1.

15:04

🧩 Advanced Quadratic Equations and Their Solutions

This part of the script addresses even more challenging equations with complex denominators. The equation is transformed into a quadratic form, and the quadratic formula is introduced as the method for solving it. The coefficients are identified, and the formula is applied to find the solutions in a specific form, a + b√13/c. The final answer is simplified to match the form requested in the question.

🔑 Final Challenge: Solving a Complex Fraction Equation

The final paragraph presents a complex equation with fractions on both sides. The common denominator is established, and the equation is simplified into a single fraction. After multiplying through by the denominators to clear the fractions, the resulting quadratic equation is solved by setting one side to zero and factoring. The solutions are found by setting each factor equal to zero, yielding x = 8.5 and x = 1/3.

Keywords

💡Algebraic Fractions

Algebraic fractions refer to expressions that represent the division of two algebraic expressions. In the context of this video, they are the core subject matter, with the script focusing on solving equations that contain these fractions. The video aims to teach viewers how to combine and simplify algebraic fractions to solve equations, as demonstrated through various examples where fractions with different denominators are combined into a single fraction with a common denominator.

💡Common Denominator

A common denominator is a shared denominator for two or more fractions, which allows for their combination into a single fraction. The script emphasizes the importance of finding the least common multiple (LCM) of the denominators to create a common denominator. This is a fundamental step in solving equations with algebraic fractions, as it simplifies the process of combining fractions on one side of the equation.

💡Lowest Common Multiple (LCM)

The lowest common multiple is the smallest number that is a multiple of each of the numbers involved. In the script, the LCM is used to determine the common denominator when combining fractions. For instance, the LCM of 6 and 4 is 12, which is then used as the common denominator to rewrite and combine the fractions, illustrating the practical application of LCM in solving algebraic equations.

💡Expanding Brackets

Expanding brackets involves distributing the terms inside the brackets across the terms outside, following the distributive property of multiplication over addition or subtraction. The script demonstrates this process as part of solving equations with algebraic fractions, where brackets are expanded to simplify the equation and facilitate further steps in the solution process.

💡Quadratic Equation

A quadratic equation is a polynomial equation of degree two, typically in the form of ax^2 + bx + c = 0. The script introduces quadratic equations as a result of simplifying and expanding algebraic fractions. The video provides examples of how to manipulate these equations to find their solutions, either by factorization or by using the quadratic formula.

💡Quadratic Formula

The quadratic formula is used to solve quadratic equations and is given by x = (-b ± √(b^2 - 4ac)) / (2a), where a, b, and c are coefficients of the quadratic equation. The script mentions the quadratic formula as a method to find the solutions for equations that do not factorize easily, showcasing its application in the context of solving more complex algebraic equations.

💡Factorization

Factorization is the process of breaking down a polynomial into a product of its factors. In the script, factorization is used as a method to simplify quadratic equations and find their solutions. The video demonstrates how certain quadratic equations can be factorized into simpler expressions, leading to the determination of the variable's value.

💡Solving Equations

Solving equations involves finding the values of the variables that make the equation true. The script's main theme revolves around solving equations with algebraic fractions. The process involves combining fractions, expanding brackets, simplifying expressions, and using various algebraic techniques to isolate the variable and find its value.

💡Like Terms

Like terms are terms in a polynomial that have the same variables raised to the same power. The script discusses collecting like terms as part of the process of simplifying equations. This simplification is crucial for combining terms and making the equation easier to solve, as demonstrated in the video's examples.

💡Completing the Square

Completing the square is a method used to solve quadratic equations by transforming the quadratic expression into a perfect square trinomial plus a constant. While the script does not explicitly use this method in the examples provided, it is mentioned as an alternative approach to solving quadratic equations that do not easily factorize.

💡Difference of Squares

The difference of squares is a special algebraic identity where the difference between two squared terms can be factored into the product of two binomials. The script mentions this concept in the context of expanding brackets and simplifying expressions, which is a useful technique for solving equations involving squared terms.

Highlights

Introduction to solving equations with algebraic fractions.

Building upon previous video on algebraic fractions operations.

Combining fractions with different denominators into one.

Finding the least common multiple (LCM) for denominators.

Rewriting fractions with a common denominator of 12.

Combining fractions over a common denominator to simplify equations.

Expanding brackets and collecting like terms to simplify equations.

Multiplying both sides by the denominator to clear fractions.

Solving a straightforward equation to find x = 27/5.

Approach to solving more complex equations with algebraic fractions.

Using the LCM of denominators 6 and 9, which is 18.

Combining fractions with a common denominator of 18.

Expanding and simplifying equations to solve for x.

Solving a quadratic equation by setting one side to zero.

Finding solutions to a quadratic equation using factorization.

Increasing difficulty with denominators that are binomial expressions.

Multiplying fractions with binomial denominators to find a common denominator.

Expanding and simplifying complex algebraic fractions.

Solving equations with quadratic results using the quadratic formula.

Using the quadratic formula to find solutions for x.

Simplifying square root expressions in the quadratic formula.

Final example with a complex equation requiring a common denominator approach.

Expanding and simplifying a complex equation with multiple terms.

Solving a quadratic equation with large coefficients.

Factorizing a quadratic equation to find possible solutions for x.

Finding two solutions for x by setting each factor equal to zero.

Conclusion and encouragement to try exam questions linked in the description.

Transcripts

[Music]

in this video we're going to learn how

to solve equations that have algebraic

fractions in them this video is going to

build upon a previous video the one

titled algebraic fractions operations

I'll put a link in this video's

description just in case you want to

check that one out

first so let's take an equation that has

some algebraic fractions in it what

we're going to do is focus on the left

hand side first of all and try and

combine these into one fraction to do

this we need to make sure we have a

common denominator at the moment the

denominators are 6 and four the lowest

common multiple of these is 12 so we

could rewrite the left hand side as two

fractions that are both over 12 if we

compare these two fractions here to get

from 6 to 12 on the bottom we must have

multiplied by two so we need to multiply

the top by two so that's 2 lots of x + 3

for the second fraction on the bottom to

get from 4 to 12 we've multiplied by

three so we multiply by three on the top

three lots of x + 1 and the right hand

side is still equal to

three now that we have the common

denominator we can combine these into

one big fraction all over 12 so we've

got two lots of x + 3 in between them we

have a plus sign and then three lots of

x + 1 and the right hand side is still

equals 3 next we need to expand the

brackets so I'm going to write the over2

and the equals 3 and then we're going to

expand the numerator here so we've got

two lots of x that's 2x two lots of

postive 3 that's POS 6 then we've got a

postive 3 * X that's POS 3x and then

pos3 * pos1 that's

pos3 we can now tidy up that numerator a

bit by collecting like terms so if we

write the over 12 and equals 3 we've got

2X and 3X that adds to make 5x and then

we've also got a six and a plus three

which gives you 9 we've now turned this

into a relatively straightforward equ

equation to solve we can multiply both

sides by 12 to get rid of that fraction

if you multiply by 12 on the left you

get 5x + 9 and if we multiply by 12 on

the right 3 * 12 is 36 then just

subtract 9 from both sides this will

give you 5x = 27 and then divide both

sides by 5 and this gives you x = 27 / 5

which is 5.4 so the solution to this

equation is

5.4 now let's try another example that's

a bit more tricky so for this one we're

going to use the same Approach at first

we're going to leave the right hand side

alone and try and combine the left hand

side into one big fraction looking at my

denominators this time of 6 and 9 the

lowest common multiple is 18 so I'm

going to write both of these as

something over 18 for the first fraction

to get from 6 to 18 we multiply by three

so we need to multiply X+ 2 on the top

by three as well so three lots of X+ 2

for the second fraction to get from 9 to

18 we multiply by two so we need two

lots of the top so two lots of x - 1 and

keep that right hand side the same equal

2/3 now we have that common denominator

we can combine them into one big

fraction all over 18 so we have three

lots of x + 2 then in between them a

subtraction sign and then two lots of x

- one and keep that right hand side the

same that's

2/3 then as we did before we're going to

expand those brackets so we'll write the

over 18 and the 2/3 down and then expand

these brackets we've got three lots of X

that's 3x three lots of pos2 that's plus

6 then we have -2 * X that's -2X and you

have to be really careful on this one

this is a really common mistake -2 * -1

is

positive2 then as before we'll collect

the like terms on the numerator so

everything else can be written down as

the same and then on the top we've got

3x - 2x that's just 1 x and then we've

got 6 + 2 which is 8 the next thing I

would do here is multiply both sides by

18 if we multiply by 18 on the left that

will cancel the 18 that's there so we

just have x + 8 and if you multiply the

right hand side by 18 we've got 2/3

multiplied by 18 that's just the same as

2/3 of 18 2/3 of 18 is 12 finally if we

subtract 8 from both sides we end up

with the answer x =

4 now let's increase the difficulty once

more and have a look at this question

here when we look at the denominators of

these fractions we can see they both

have two terms this means to find their

lowest common multiple we're going to

need to multiply them together so both

of these fractions can be Rewritten as

something over x + 2 x + 5 let's look at

the left fraction first so this one to

get from x + 2 to x + 2 * x + 5 we

obviously multiply by x + 5 so we need

to do this on the top as well so we need

four lots of x + 5 and then we can look

at the second fraction so to get from x

+ 5 to x + 2 * x + 5 we multiply by x +

2 so on the top we need to multiply by

x+2 as well and that's four lots of x+2

and of course the right hand side is

still just

two now that we have this common

denominator we can write them as one big

fraction all over x + 2 x + 5 so on the

numerators we have four lots of X+ 5

then in between them we have a plus sign

and then four lots of X+ 2 and the right

hand side is still equals 2 we can now

do some expanding and then simplifying

so if we expand out the numerator we've

got 4 * X which is 4X 4 * 5 which is +

20 4 * X which is 4X again and then 4 *

2 that's a positive

8 next we need to do the simplifying of

the numerator so on the top here we have

4x + 4x which is 8 x and then we also

have 20 + 8 which is

28 next we're going to multiply both

sides of this equation by both of those

brackets that are on that denominator so

by X+ 2 and x + 5 this will clear the

fraction on the left hand side to give

us 8x +

28 but on the the right hand side we'll

have two lots of x + 2 x + 5 next we'll

leave the left hand side alone and then

we've got two lots and then we'll expand

those brackets and whatever we get will

go inside this much larger bracket here

so if we expanded that double bracket

we'd have x * X that's x 2 we then have

x * 5 which is 5 x and 2 * X which is 2

X and 5 x and 2 x adds to make 7 x and

then finally we'd have 2 * 5 which is 10

on the right hand side we can then

expand this bra ET so if we leave the

left hand side alone on the right hand

side we have two lots of everything

that's in that bracket so two lots of

x^2 two lots of 7 x that's 14x two lot

of 10 that's 20 you can now see we have

a quadratic equation to solve when we

solve these we need to set one of the

sides equal to zero since the 2x^2 is on

the right hand side I'm going to make

the left hand side equal to zero to do

this I need to take away 8X and also

take away 28 from both sides if I took

away those from the left hand side I'd

get zero and if if I take those away

from the right hand side the 2x^2 will

remain unchanged but then I need to take

the 8X away from the 14x which is 6X and

then take the 28 away from the 20 which

is8 now there's a common factor of two

here to all terms so I can divide both

sides by two if I divide the left hand

side by two 0id by 2 is still 0o but

then on the right hand side 2x^2 ID 2 is

1 x^2 6X / 2 is 3x and8 ID 2 is -4

fortunately for us this quadratic will

factorize so on the left hand side we

have zero and the right hand side will

factorize to give x + 4 x -1 this leads

us to two solutions either the first

bracket is zero so x + 4 equal 0 in

which case X is -4 or the second bracket

equals 0 so X - 1 = 0 gives us the

solution X =

pos1 now let's have a look at an even

trickier question so for this one we

need to solve an equation but we've told

we need to give the answer in a

particular form

a plus orus < TK 13 / B this is a clear

indication that when we get to a

quadratic at the end it's not going to

factorize anymore we either need to use

the quadratic formula or complete the

square to get the solutions so let's go

ahead and try and solve this one so

first of all we'll look at those

denominators and we can see we've got x

+ 3 x - 3 so the product of those will

become our

denominators then we compare the first

fractions so to get from x + 3 to x + 3

x - 3 we multiply by x - 3 so we need to

multiply that six on the top by x - 3

and when we compare the second fractions

this time we've multiplied by x + 3 so

on the top we need two lots of x + 3 and

this equals 3 now we can combine them

into one big fraction over that common

denominator we have 6 lots of x - 3 in

between the fractions is a subtraction

sign and then 2 lots of x +

3 now we can expand out those brackets

so on the top we've got 6 * X that's 6 x

6 * -3

8 -2 * x -2X and a -2 * a POS 3 that's

A6 then we can collect those like terms

so on the top we've got 6X take 2x

that's 4X and then -8 subtract 6 that's

-4 then we would multiply both sides by

that denominator x + 3 x - 3 this will

clear the fraction on the left hand side

so you've just got 4x - 24 but on the

right hand side we have three lots of x

+ 3 x - 3 to solve this we'll leave the

left hand side alone and on the right

hand side we've got three lots of

whatever we get when we expand these two

brackets well this one's the difference

of two squares if we do x * X we get x2

and then we have x * -3 -3x but also

posi 3 * X which is postive 3x so those

X terms will cancel and then we have 3 *

3 which is 9 so we have 4x - 24 and then

if we expand this bracket three L of x^2

is 3x^2 and 3 lot of -9 is

-27 so you can see we do have a

quadratic and we need to get one of the

sides to be equal to zero once again I'm

going to make that the left hand side so

to make the left hand side equal to zero

I would need to subtract 4X and then add

24 to both sides so if I do that on the

left I get zero of course and on the

right hand side I get 3x^2 that remains

unchanged then I subtracted 4X and

there's no X term there already so

that's just -4x and then we need to add

24 to -27 which gives us -3 so we end up

with this quadratic equation here to

solve now remember we know from the form

they want the answer in we need to solve

this using the formula or completing the

square and this one's probably going to

be much easier using the quadratic

formula so we need to find the values of

a b and c they're the coefficients of X2

X and the constant term so the

coefficient of X2 is 3 the coefficient

of x is -4 and the constant term at the

end is -3 so a is 3 B is-4 and C is -3

now we can substitute those values into

the quadratic formula it's x equal

negative B well B is already negative so

if we did negative B it will become

positive so it's actually a positive 4

plus or minus the square root of then

b^2 so -4 2 subtract four lots of a

which is 3 multipli C which is -3 and

all of this is divided by two lots of a

a and 2 lot of a is 2 lots of three so

six now this question would actually be

a non-calculator question so we're going

to need to work out the value of the

number that's inside that square root so

let's write x = 4 plus or minus the < TK

of over 6 but we'll work out what's

inside that square root so if we did -4

squar that would be -4 multiplied by

itself so

16 and then if we did4 multipli by 3

that's -12 and then multiply that by -3

that becomes a positive

36 so we end up with 16 + 36 inside that

square root you can add those two

together that's 52 so x = 4 plus or- <

TK 52 / 6 next what we need to do is

simplify that < TK 52 so using our third

rules if we take root 52 that could be

expressed Asun 4 multiplied

byun3 the square < TK 4 is just two so

this is 2 < tk13 so in place ofun 52 we

could write 2K3 so it's X = 4 plus orus

2un 13 now over 6 now this is looking a

lot like the answer but we're still not

quite there that one just has 1 < TK 13

and we have two the reason for that is

we have a common factor of two here if

we divide all terms by two we get x

equals if we divide the four by two we

get two divide 2 < tk3 by two you get

1un 13 and divide 6 by two you get three

this answer now matches the form we were

asked in the question you can see the

value of a is 2 2 and the value of B is

three now we're going to look at one

more final question this one here we're

going to start this one how we started

all of the previous questions we're

going to write the left hand side over a

common denominator so for this one it'll

be the product of x + 17 and 2x -

5 so if we compare these first two

fractions you can see we've multiplied

by 2x - 5 so if we multiply by 2x - 5 on

the top we get 3 x lots of 2x -

5 and and for the right fractions when

we compare these we've multiplied by x +

17 so we need three lots of x + 17 and

the right hand side is equal to 3 over

4 now we have that common denominator we

can write them as one big fraction so

we've got 3x and then brackets 2x - 5

then we have a subtraction in between

them and 3x +

17 next we're going to expand out that

numerator so we'll write everything else

the same we have 3x * 2x that's 6 x^2 3x

* -5 that's

15x -3 * X -3x and -3 * pos7 is

-51 now there's a little bit of

simplifying we can do on the numerator

so we can't simplify the 6x2 that's the

only term of x^2 but we can simplify -5x

subtract 3x which is -8x and then

subtract

51 so now we've written the left hand

side as a single fraction what we're

going to do next is multiply both sides

by both of those brackets on that

denominator so if we multiply the left

hand side by x + 17 and 2x - 5 that will

clear the fraction so we've got 6x^2 -

18x - 51 and on the right hand side

we've just multiplied this

3/4 by both of those brackets now in the

previous couple of questions we haven't

had to deal with a fraction like this on

the right hand side it was just an

integer before those two brackets to

deal with this fraction you can just

multiply both sides by four if we

multiply the left hand side by four we

get four lots of the left hand side and

if we multiply the right hand side by

four that four will cancel so we've just

got three lots of those two

brackets now we've got lots of expanding

to do so on the left hand side we need

to multiply by four so we've got four

lots of 6x^2 that's 24 x^2 four lots of

-8x that's - 72x and four Lots of- 51 is

-24 then on the right hand side we're

going to write a three and then a big

bracket then expand out the double

bracket and write the terms inside this

big bracket ET so we've got x * 2x first

that's

2x^2 then we would do x * -5 so -5x but

we would also do 17 * 2x which is a

positive

34x so we've got - 5x + 34x that's

positive

29x and finally 17 * -5 is

85 we can leave that left hand side

alone and expand this bracket on the

right hand side we need to multiply by

three this time 3 lots of 2x^2 is 6x^2 3

lots of 29x is 87x and 3 lots of 85 is -

255 so we do end up with a quadratic

equation to solve since we have more X

squs on the left hand side I'm going to

make the right hand side equal to zero

on this one so i' need to subtract 6x^2

subtract 87x and add 255 to both sides

so on the left hand side if I subtract 6

x^2 from 24 x^2 I get 18 x^2 and then if

I subtract 87x from the 72x we get

159x and then if we add 255 to ne24 we

end up with a positive 51 and the right

hand side we know will equal

zero now let's take that quadratic

equation and we're going to solve it

this one does look really nasty because

of the size of the numbers but there is

actually a common factor of three here

if you divide 18 x^2 by 3 you get 6 x^2

divide -9x by 3 that's - 53x and divide

the 51 by 3 that's 17 and of course if

you divide 0 by three that's still zero

so we're going to factorize this one

actually it will factorize into two

brackets now it's actually quite

fortunate we have a 17 at the end since

we have a 17 that's a prime number we

know its factors are only 1 and 17 so if

it did factorize the only possible

numbers we could put at the end of these

brackets would be 1 and 17 and they

would actually both be negative as well

since a negative times a negative gives

you a positive so we still get a

positive 17 but the only way we'd end up

with a 53x is if we had negative terms

here if they were both positive that 53x

wouldn't be attainable so we can put 17

and Ne -1 in these brackets and then we

just need to think about a few of the

different combinations to try and get

that 6x^ s so we're either going to use

a 6X and a 1x or a 3X and a 2x now if

you try a few different combinations out

it shouldn't take you too long to

realize we need 2x - 17 and 3x - 1 if

you won't give this a go and expand it

out and you'll see it does give that

quadratic a b

so finally to get our two solutions we

need to set the first bracket equal to 0

so 2x - 17 = 0 or the second bracket

equal to 0 so 3x - 1 = 0 to solve the

left one you can add 17 to both sides

and get 2x = 17 and then divide both

sides by two to get x = 8.5 for the

right one you can add one to both sides

so 3x = 1 and divide both sides by 3 to

get x =

1/3 thank you for watching this video I

hope you found a useful check out the

one I think you should watch next

subscribe so you don't miss out on my

future videos and now go ahead and try

the exam questions I've Linked In this

video's

description