Answer:
46
Step-by-step explanation:
5x = 3x - 2 + 34
5x = 3x + 32
2x = 32
x = 16
plug back in
3( 16 ) - 2
48 - 2
46
<u>Question 8</u>
a^2 + 7a + 12
= (a+3)(a+4)
When factorising a quadratic, the product of the two factors should equal the constant term (12), and the sum of the two factors should equal the linear term (7). To find the two factors, list out the factors of 12 (1x12, 2x6, 3x4) and identify the pair that adds up to 7 (3+4).
An alternative method if you get stuck during your exam would be to solve it algebraically using the quadratic formula and then write it in the factorised form.
a = (-7 +or- sqrt(7^2 - 4(1)(12)) / 2(1)
= (-7 +or- sqrt(1))/2
= -3 or -4
These factors are the negative of the values that would go in the brackets when written in factorised form, as when a = -3 the factor (a+3) would equal 0. (If it were positive 3 instead, then in the factorised form it would be a-3).
<u>Question 10</u>
-3(x - y)/9 + (4x - 7y)/2 - (x + y)/18
Rewrite each fraction with a common denominator so you can combine the fractions into one.
= -6(x - y)/18 + 9(4x - 7y)/18 - (x + y)/18
= (-6(x - y) + 9(4x - 7y) - (x + y)) /18
Expand the brackets and collect like terms.
= (-6x + 6y + 36x - 63y - x - y)/18
= (29x - 58y)/18
= 29/18 x - 29/9 y
Answer: The system of equations is:
x + 2y + 2 = 4
y - 3z = 9
z = - 2
The solution is: x = -22; y = 15; z = -2;
Step-by-step explanation: ONe way of solving a system of equations is using the Gauss-Jordan Elimination.
The method consists in transforming the system into an augmented matrix, which is writing the system in form of a matrix and then into a <u>Row</u> <u>Echelon</u> <u>Form,</u> which satisfies the following conditions:
- There is a row of all zeros at the bottom of the matrix;
- The first non-zero element of any row is 1, which called leading role;
- The leading row of the first row is to the right of the leading role of the previous row;
For this question, the matrix is a Row Echelon Form and is written as:
![\left[\begin{array}{ccc}1&2&2\\0&1&3\\0&0&1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%262%262%5C%5C0%261%263%5C%5C0%260%261%5Cend%7Barray%7D%5Cright%5D)
![\left[\begin{array}{ccc}4\\9\\-2\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D4%5C%5C9%5C%5C-2%5Cend%7Barray%7D%5Cright%5D)
or in system form:
x + 2y + 2z = 4
y + 3z = 9
z = -2
Now, to determine the variables:
z = -2
y + 3(-2) = 9
y = 15
x + 30 - 4 = 4
x = - 22
The solution is (-22,15,-2).
The formula for the discriminant is b^2-4ac. The formula you wote is in the form of
ax^2+c=0, so first, you need to bring that -2 to the left. When you do that, you get (the original equation would be ax^2+bx+c but you have no bx.)
3x^2-8
So, since there's no b, it would be
b^2-4ac
0^2-4(3)(-8
-12 x -8
96
