Math / Algebra

Question(suite)
7 Trouve la valeur des expressions algébriques suivantes. a) $\left(2 x^{2}-4 x+3\right) \div 3$ , si $x=9$ b) $8 a b+11 b-30$ , si $a=2$ et $b=-3$

Studdy Solution

STEP 1

What is this asking? We need to substitute the given values of $x$ , $a$ , and $b$ into two algebraic expressions and calculate the results! Watch out! Be super careful with those negative signs and the order of operations (PEMDAS/BODMAS)!

STEP 2

1. Solve for expression (a)
2. Solve for expression (b)

STEP 3

We are given the expression $\left(2x^{2} - 4x + 3\right) \div 3$ and we're told that $x = 9$ .
Let's plug in that value of $x$ !
So, we get $\left(2 \cdot 9^{2} - 4 \cdot 9 + 3\right) \div 3$ .

STEP 4

Remember PEMDAS/BODMAS!
Exponents come before multiplication, division, addition, or subtraction.
So, $9^2$ is $9 \cdot 9 = \textbf{81}$ .
Our expression becomes $\left(2 \cdot \textbf{81} - 4 \cdot 9 + 3\right) \div 3$ .

STEP 5

Next up in PEMDAS/BODMAS is multiplication.
We have two multiplications: $2 \cdot 81 = \textbf{162}$ and $4 \cdot 9 = \textbf{36}$ .
Now our expression looks like this: $\left(\textbf{162} - \textbf{36} + 3\right) \div 3$ .

STEP 6

Inside the parentheses, we subtract: $162 - 36 = \textbf{126}$ .
Then, we add: $\textbf{126} + 3 = \textbf{129}$ .
Our expression simplifies to $\textbf{129} \div 3$ .

STEP 7

Finally, we divide $\textbf{129}$ by $3$ . $129 \div 3 = \textbf{43}$ .
Awesome!

STEP 8

We're given the expression $8ab + 11b - 30$ and we know that $a = 2$ and $b = -3$ .
Let's substitute those values: $8 \cdot 2 \cdot (-3) + 11 \cdot (-3) - 30$ .

STEP 9

First, $8 \cdot 2 = \textbf{16}$ .
Then, $\textbf{16} \cdot (-3) = \textbf{-48}$ .
Next, $11 \cdot (-3) = \textbf{-33}$ .
Our expression becomes $\textbf{-48} + \textbf{(-33)} - 30$ .

STEP 10

Remember, adding a negative number is the same as subtracting.
So, $-48 + (-33)$ is the same as $-48 - 33 = \textbf{-81}$ .
Then, $\textbf{-81} - 30 = \textbf{-111}$ .

STEP 11

a) The value of $\left(2x^{2} - 4x + 3\right) \div 3$ when $x = 9$ is $\textbf{43}$ . b) The value of $8ab + 11b - 30$ when $a = 2$ and $b = -3$ is $\textbf{-111}$ .

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Question(suite)
7 Trouve la valeur des expressions algébriques suivantes. a) $\left(2 x^{2}-4 x+3\right) \div 3$ , si $x=9$ b) $8 a b+11 b-30$ , si $a=2$ et $b=-3$

Studdy Solution

STEP 1

What is this asking? We need to substitute the given values of $x$ , $a$ , and $b$ into two algebraic expressions and calculate the results! Watch out! Be super careful with those negative signs and the order of operations (PEMDAS/BODMAS)!

STEP 2

1. Solve for expression (a)
2. Solve for expression (b)

STEP 3

We are given the expression $\left(2x^{2} - 4x + 3\right) \div 3$ and we're told that $x = 9$ .
Let's plug in that value of $x$ !
So, we get $\left(2 \cdot 9^{2} - 4 \cdot 9 + 3\right) \div 3$ .

STEP 4

Remember PEMDAS/BODMAS!
Exponents come before multiplication, division, addition, or subtraction.
So, $9^2$ is $9 \cdot 9 = \textbf{81}$ .
Our expression becomes $\left(2 \cdot \textbf{81} - 4 \cdot 9 + 3\right) \div 3$ .

STEP 5

Next up in PEMDAS/BODMAS is multiplication.
We have two multiplications: $2 \cdot 81 = \textbf{162}$ and $4 \cdot 9 = \textbf{36}$ .
Now our expression looks like this: $\left(\textbf{162} - \textbf{36} + 3\right) \div 3$ .

STEP 6

Inside the parentheses, we subtract: $162 - 36 = \textbf{126}$ .
Then, we add: $\textbf{126} + 3 = \textbf{129}$ .
Our expression simplifies to $\textbf{129} \div 3$ .

STEP 7

Finally, we divide $\textbf{129}$ by $3$ . $129 \div 3 = \textbf{43}$ .
Awesome!

STEP 8

We're given the expression $8ab + 11b - 30$ and we know that $a = 2$ and $b = -3$ .
Let's substitute those values: $8 \cdot 2 \cdot (-3) + 11 \cdot (-3) - 30$ .

STEP 9

First, $8 \cdot 2 = \textbf{16}$ .
Then, $\textbf{16} \cdot (-3) = \textbf{-48}$ .
Next, $11 \cdot (-3) = \textbf{-33}$ .
Our expression becomes $\textbf{-48} + \textbf{(-33)} - 30$ .

STEP 10

Remember, adding a negative number is the same as subtracting.
So, $-48 + (-33)$ is the same as $-48 - 33 = \textbf{-81}$ .
Then, $\textbf{-81} - 30 = \textbf{-111}$ .

STEP 11

a) The value of $\left(2x^{2} - 4x + 3\right) \div 3$ when $x = 9$ is $\textbf{43}$ . b) The value of $8ab + 11b - 30$ when $a = 2$ and $b = -3$ is $\textbf{-111}$ .

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Question(suite)7 Trouve la valeur des expressions algébriques suivantes. a) (2x2−4x+3)÷3\left(2 x^{2}-4 x+3\right) \div 3(2x2−4x+3)÷3, si x=9x=9x=9 b) 8ab+11b−308 a b+11 b-308ab+11b−30, si a=2a=2a=2 et b=−3b=-3b=−3

Studdy Solution

STEP 1

What is this asking? We need to substitute the given values of xxx, aaa, and bbb into two algebraic expressions and calculate the results! Watch out! Be super careful with those negative signs and the order of operations (PEMDAS/BODMAS)!

STEP 2

1. Solve for expression (a)2. Solve for expression (b)

STEP 3

We are given the expression (2x2−4x+3)÷3\left(2x^{2} - 4x + 3\right) \div 3(2x2−4x+3)÷3 and we're told that x=9x = 9x=9.Let's **plug in** that value of xxx!So, we get (2⋅92−4⋅9+3)÷3\left(2 \cdot 9^{2} - 4 \cdot 9 + 3\right) \div 3(2⋅92−4⋅9+3)÷3.

STEP 4

Remember PEMDAS/BODMAS!Exponents come before multiplication, division, addition, or subtraction.So, 929^292 is 9⋅9=819 \cdot 9 = \textbf{81}9⋅9=81.Our expression becomes (2⋅81−4⋅9+3)÷3\left(2 \cdot \textbf{81} - 4 \cdot 9 + 3\right) \div 3(2⋅81−4⋅9+3)÷3.

STEP 5

Next up in PEMDAS/BODMAS is multiplication.We have two multiplications: 2⋅81=1622 \cdot 81 = \textbf{162}2⋅81=162 and 4⋅9=364 \cdot 9 = \textbf{36}4⋅9=36.Now our expression looks like this: (162−36+3)÷3\left(\textbf{162} - \textbf{36} + 3\right) \div 3(162−36+3)÷3.

STEP 6

Inside the parentheses, we subtract: 162−36=126162 - 36 = \textbf{126}162−36=126.Then, we add: 126+3=129\textbf{126} + 3 = \textbf{129}126+3=129.Our expression simplifies to 129÷3\textbf{129} \div 3129÷3.

STEP 7

Finally, we divide 129\textbf{129}129 by 333. 129÷3=43129 \div 3 = \textbf{43}129÷3=43.Awesome!

STEP 8

We're given the expression 8ab+11b−308ab + 11b - 308ab+11b−30 and we know that a=2a = 2a=2 and b=−3b = -3b=−3.Let's **substitute** those values: 8⋅2⋅(−3)+11⋅(−3)−308 \cdot 2 \cdot (-3) + 11 \cdot (-3) - 308⋅2⋅(−3)+11⋅(−3)−30.

STEP 9

First, 8⋅2=168 \cdot 2 = \textbf{16}8⋅2=16.Then, 16⋅(−3)=-48\textbf{16} \cdot (-3) = \textbf{-48}16⋅(−3)=-48.Next, 11⋅(−3)=-3311 \cdot (-3) = \textbf{-33}11⋅(−3)=-33.Our expression becomes -48+(-33)−30\textbf{-48} + \textbf{(-33)} - 30-48+(-33)−30.

STEP 10

Remember, adding a negative number is the same as subtracting.So, −48+(−33)-48 + (-33)−48+(−33) is the same as −48−33=-81-48 - 33 = \textbf{-81}−48−33=-81.Then, -81−30=-111\textbf{-81} - 30 = \textbf{-111}-81−30=-111.

STEP 11

a) The value of (2x2−4x+3)÷3\left(2x^{2} - 4x + 3\right) \div 3(2x2−4x+3)÷3 when x=9x = 9x=9 is 43\textbf{43}43. b) The value of 8ab+11b−308ab + 11b - 308ab+11b−30 when a=2a = 2a=2 and b=−3b = -3b=−3 is -111\textbf{-111}-111.

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Question(suite)
7 Trouve la valeur des expressions algébriques suivantes. a) $\left(2 x^{2}-4 x+3\right) \div 3$ , si $x=9$ b) $8 a b+11 b-30$ , si $a=2$ et $b=-3$

What is this asking? We need to substitute the given values of $x$ , $a$ , and $b$ into two algebraic expressions and calculate the results! Watch out! Be super careful with those negative signs and the order of operations (PEMDAS/BODMAS)!

1. Solve for expression (a)
2. Solve for expression (b)

We are given the expression $\left(2x^{2} - 4x + 3\right) \div 3$ and we're told that $x = 9$ .
Let's plug in that value of $x$ !
So, we get $\left(2 \cdot 9^{2} - 4 \cdot 9 + 3\right) \div 3$ .

Remember PEMDAS/BODMAS!
Exponents come before multiplication, division, addition, or subtraction.
So, $9^2$ is $9 \cdot 9 = \textbf{81}$ .
Our expression becomes $\left(2 \cdot \textbf{81} - 4 \cdot 9 + 3\right) \div 3$ .

Next up in PEMDAS/BODMAS is multiplication.
We have two multiplications: $2 \cdot 81 = \textbf{162}$ and $4 \cdot 9 = \textbf{36}$ .
Now our expression looks like this: $\left(\textbf{162} - \textbf{36} + 3\right) \div 3$ .

Inside the parentheses, we subtract: $162 - 36 = \textbf{126}$ .
Then, we add: $\textbf{126} + 3 = \textbf{129}$ .
Our expression simplifies to $\textbf{129} \div 3$ .

Finally, we divide $\textbf{129}$ by $3$ . $129 \div 3 = \textbf{43}$ .
Awesome!

We're given the expression $8ab + 11b - 30$ and we know that $a = 2$ and $b = -3$ .
Let's substitute those values: $8 \cdot 2 \cdot (-3) + 11 \cdot (-3) - 30$ .

First, $8 \cdot 2 = \textbf{16}$ .
Then, $\textbf{16} \cdot (-3) = \textbf{-48}$ .
Next, $11 \cdot (-3) = \textbf{-33}$ .
Our expression becomes $\textbf{-48} + \textbf{(-33)} - 30$ .

Remember, adding a negative number is the same as subtracting.
So, $-48 + (-33)$ is the same as $-48 - 33 = \textbf{-81}$ .
Then, $\textbf{-81} - 30 = \textbf{-111}$ .

a) The value of $\left(2x^{2} - 4x + 3\right) \div 3$ when $x = 9$ is $\textbf{43}$ . b) The value of $8ab + 11b - 30$ when $a = 2$ and $b = -3$ is $\textbf{-111}$ .