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3 years ago

What is the solution to this system of linear equations 3x-2y=14 5x+y=32 brainly

Mathematics

2 answers:

den301095 [7]3 years ago

8 0

$3x-2y=14\\5x+y=32$

Multiply 2 in the 5x+y=32 so that we can eliminate the "y"

$3x-2y=14\\5x(2)+(2)y=32(2)$

From 5x(2)+(2)y=32(2), $10x+2y=64$

$3x-2y=14\\10x+2y=64$

Then start eliminating the y,

$13x=78$ [3x+10x = 13x] [-2y+2y = 0] [14+64 = 78]

$x=\frac{78}{13}$

Therefore, $x=6$ but we are not done yet, we need to find the value of y.

Substitute x = 6 in any equations but only one equation so I'll substitute in 5x+y=32 instead.

$5(6)+y=32\\30+y=32\\y=32-30\\y=2$

Therefore, y = 2.

So the answer is x = 6, y = 2 or (6,2)

xxTIMURxx [149]3 years ago

7 0

Answer:

x = 6

y = 2

Step-by-step explanation:

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Step 1: We make the assumption that 498 is 100% since it is our output value.

Step 2: We next represent the value we seek with $x$x.

Step 3: From step 1, it follows that $100\%=498$100%=498.

Step 4: In the same vein, $x\%=4$x%=4.

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Please help me with this question I also need to see the steps if possible!

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Hi there!

To solve, we must use the following trig identity:

sin(u - v) = sin(u)cos(v) - sin(v)cos(u)

We can rewrite the left hand side of the equation as:

$\frac{sin(u)cos(v)-sin(v)cos(u)}{sin(u)cos(v)}$

Split the fraction:

$\frac{sin(u)cos(v)}{sin(u)cos(v)} - \frac{sin(v)cos(u)}{sin(u)cos(v)} =$

First fraction reduces to 1:

$1 - \frac{sin(v)cos(u)}{sin(u)cos(v)} =$

Simpify each with common arguments:

$1 - tan(v)cot(u)$

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3 years ago