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3 years ago

Y – 5 = x 4x – y = 4 elimination

Mathematics

2 answers:

Alex787 [66]3 years ago

6 0

Answer:

(3,8)

Step-by-step explanation:

y - 5 = x

4x - y = 4

Since y-5 = x, substitute y-5 in place of x in the second equation:

4x- y = 4

4(y-5) - y = 4

4y - 4*5 - y = 4

3y - 20 = 4

3y = 24

y =8

Solve for x:

y - 5 = x

8-5 = x

3 = x

Check:

4x - y = 4

4(3) - 8 = 4

12 - 8 = 4

4 = 4

Check

y - 5 = x

8 - 5 = 3

3 = 3

Check!

jasenka [17]3 years ago

4 0

y - 5 = x

4x - y = 4

Arrange:

y - 5 = x

-x + y = 5

-x + y = 5

4x - y = 4

Add together:

-x + y = 5

4x - y = 4

3x = 9

Solve for x:

x = 9 ÷ 3

x = 3

Substitute and solve y:

y - 5 = x

y - 5 = 3

y = 3 + 5

y = 8

x = 3

y = 8

Read more about triangles at

brainly.com/question/17335144

#SPJ1

5 0

1 year ago

Sam takes out a $167,000 mortgage for 20 years. He makes monthly payments and at the end calculates

cupoosta [38]

Answer:

The Annual rate of interest for the mortgage is 1.8%

Step-by-step explanation:

Given as :

The mortgage principal = p = $167,000

The time period of mortgage = t = 20 years

The Amount paid towards mortgage in 20 years = A = $240,141

Let the Annual percentage rate on interest = r % compounded annually

Now, <u>From Compound Interest method</u>

Amount = Principal × $(1+\dfrac{\textrm rate}{100})^{\textrm time}$

Or, A = p × $(1+\dfrac{\textrm r}{100})^{\textrm t}$

Or, $240,141 = $167,000 × $(1+\dfrac{\textrm r}{100})^{\textrm 20}$

or, $\dfrac{240,141}{167,000}$ = $(1+\dfrac{\textrm r}{100})^{\textrm 20}$

Or , 1.437 = $(1+\dfrac{\textrm r}{100})^{\textrm 20}$

Or, $(1.437)^{\frac{1}{20}}$ = $(1+\dfrac{r}{100})$

or, 1.018 = $(1+\dfrac{r}{100})$

Or, $\dfrac{r}{100}$ = 1.018 - 1

Or, $\dfrac{r}{100}$ = 0.018

∴ r = 0.018 × 100

i.e r = 1.8

So, The rate of interest applied = r = 1.8 %

Hence, The Annual rate of interest for the mortgage is 1.8% Answer

8 0

3 years ago

How to find matrics M

makkiz [27]

$\left[\begin{array}{ccc}9&7\\-4&2\end{array}\right] -4M=\left[\begin{array}{ccc}-3&3\\4&-18\end{array}\right]\qquad(*) \\\\M=\left[\begin{array}{ccc}a&b\\c&d\end{array}\right]\to4M=\left[\begin{array}{ccc}4a&4b\\4c&4d\end{array}\right]$

$(*)\qquad\left[\begin{array}{ccc}9&7\\-4&2\end{array}\right] -\left[\begin{array}{ccc}4a&4b\\4c&4d\end{array}\right]=\left[\begin{array}{ccc}-3&3\\4&-18\end{array}\right]\\\\\left[\begin{array}{ccc}9-4a&7-4b\\-4-4c&2-4d\end{array}\right]=\left[\begin{array}{ccc}-3&3\\4&-18\end{array}\right]\\.\qquad\qquad\qquad\qquad\ \ \ \Downarrow\\(_{11})\qquad9-4a=-3\ \ \ |-9\\-4a=-12\ \ \ \ |:(-4)\\a=3\\\\(_{12})\qquad7-4b=3\ \ \ \ |-7\\-4b=-4\ \ \ \ |:(-4)\\b=1$

$(_{21})\qquad-4-4c=4\ \ \ \ |+4\\-4c=8\ \ \ \ |:(-4)\\c=-2\\\\(_{22})\qquad2-4d=-18\ \ \ \ |-2\\-4d=-20\ \ \ \ \ |:(-4)\\d=5\\\\Answer:\ A.\ M=\left[\begin{array}{ccc}3&1\\-2&5\end{array}\right]$