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4 years ago

Year 9 Maths - linear equations please help

Mathematics

2 answers:

MAXImum [283]4 years ago

6 0

X=3, Y=9
X+Y=12
Y=X+6

X+X+6=12
2X=6
X=3

Y=3+6=9

jek_recluse [69]4 years ago

3 0

Answer:

x=1 & y=6 or (1,6)

Step-by-step explanation:

3x-x+5=7 Substitute the 2nd equation in for y

2x+5=7 Do 3x-x to get 2x

2x=2 Subtract 5 from both sides

x=1 Divide 2 from both sides

y=1+5 Insert 1 as x

y=6 Add 1+5 to solve for y

Hope this helped!

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Y(2y-3)(5y-1) check using foil

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3 years ago

For the following linear system, put the augmented coefficient matrix into reduced row-echelon form.

Anni [7]

Answer:

The reduced row-echelon form of the linear system is $\left[\begin{array}{cccc}1&0&-5&0\\0&1&3&0\\0&0&0&1\end{array}\right]$

Step-by-step explanation:

We will solve the original system of linear equations by performing a sequence of the following elementary row operations on the augmented matrix:

Interchange two rows
Multiply one row by a nonzero number
Add a multiple of one row to a different row

To find the reduced row-echelon form of this augmented matrix

$\left[\begin{array}{cccc}2&3&-1&14\\1&2&1&4\\5&9&2&7\end{array}\right]$

You need to follow these steps:

Divide row 1 by 2 $\left(R_1=\frac{R_1}{2}\right)$

$\left[\begin{array}{cccc}1&3/2&-1/2&7\\1&2&1&4\\5&9&2&7\end{array}\right]$

Subtract row 1 from row 2 $\left(R_2=R_2-R_1\right)$

$\left[\begin{array}{cccc}1&3/2&-1/2&7\\0&1/2&3/2&-3\\5&9&2&7\end{array}\right]$

Subtract row 1 multiplied by 5 from row 3 $\left(R_3=R_3-\left(5\right)R_1\right)$

$\left[\begin{array}{cccc}1&3/2&-1/2&7\\0&1/2&3/2&-3\\0&3/9&9/2&-28\end{array}\right]$

Subtract row 2 multiplied by 3 from row 1 $\left(R_1=R_1-\left(3\right)R_2\right)$

$\left[\begin{array}{cccc}1&0&-5&16\\0&1/2&3/2&-3\\0&3/9&9/2&-28\end{array}\right]$

Subtract row 2 multiplied by 3 from row 3 $\left(R_3=R_3-\left(3\right)R_2\right)$

$\left[\begin{array}{cccc}1&0&-5&16\\0&1/2&3/2&-3\\0&0&0&-19\end{array}\right]$

Multiply row 2 by 2 $\left(R_2=\left(2\right)R_2\right)$

$\left[\begin{array}{cccc}1&0&-5&16\\0&2&3&-6\\0&0&0&-19\end{array}\right]$

Divide row 3 by −19 $\left(R_3=\frac{R_3}{-19}\right)$

$\left[\begin{array}{cccc}1&0&-5&16\\0&2&3&-6\\0&0&0&1\end{array}\right]$

Subtract row 3 multiplied by 16 from row 1 $\left(R_1=R_1-\left(16\right)R_3\right)$

$\left[\begin{array}{cccc}1&0&-5&0\\0&1&3&-6\\0&0&0&1\end{array}\right]$

Add row 3 multiplied by 6 to row 2 $\left(R_2=R_2+\left(6\right)R_3\right)$

$\left[\begin{array}{cccc}1&0&-5&0\\0&1&3&0\\0&0&0&1\end{array}\right]$

8 0

3 years ago

I know the answer I got is wrong help.

irinina [24]

Answer:

∠2 = 18°

Step-by-step explanation:

∠WXZ = ∠1 + ∠2 ← substitute ∠1 = 3∠2

∠WXZ = 3∠2 + ∠2, that is

72 = 4∠2 ( divide both sides by 4 )

18 = ∠2, that is

∠2 = 18°

8 0

4 years ago

I need help on number 6

sergeinik [125]

D is correct
3(2x-5)=6x+k
6x-15=6x+k
If 2 equations are identical the solution will be infinite

4 0

3 years ago

I need help solving these problems please.

ddd [48]

5. 15 x 3/4 = 11.25

6. 10 x 1/2 = 5

7. 11 x 2 = 22

8. 12/1.5 = 8

5 0

4 years ago