What is the y-coordinate of the solution for the system of equations?

Mathematics

2 answers:

enot [183]4 years ago

7 0

Y= 12 - x

27 + 3( 12 - x ) = 2x

KATRIN_1 [288]4 years ago

7 0

First equation is: x - y = 12
x = 12 + y

Now, substitute this in 2nd equation,
2x - 3y = 27
2(12+y) - 3y = 27
24 + 2y - 3y = 27
-y = 27 - 24
y = -3

In short, Your Answer would be -3

Hope this helps!

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o-na [289]

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$\blue\textsf{\textbf{Question:-}}}}$

Solve 3x+4=28

$\blue\textsf{\textbf{\underline{\underline{Answer and how to solve:-}}}}$

First, subtract 4 on both sides:-

$\tt{3x=28-4}$

$\tt{3x=24}$

Divide by 3 on both sides:-

$\dashrightarrow\bigstar\boxed{\boxed{\bold{x=8}}}\dashleftarrow$

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8 0

2 years ago

Read 2 more answers

If 1/√a-√b=1/3 and 1/√a+√b=1/2, then find the difference of a and b.

kow [346]

<u>ANSWER:</u>

If $\frac{1}{\sqrt{a}-\sqrt{b}}=\frac{1}{3}$ and $\frac{1}{\sqrt{a}+\sqrt{b}}=\frac{1}{2}$ then the difference of a and b is 6

<u>SOLUTION:</u>

Given, $\frac{1}{\sqrt{a}-\sqrt{b}}=\frac{1}{3}$ → $\sqrt{a}-\sqrt{b}=3$ ----- (1)

And $\frac{1}{\sqrt{a}+\sqrt{b}}=\frac{1}{2}$ → $\sqrt{a}+\sqrt{b}=2$ --- (2)

We have to find difference of a and b.

Now, add (1) and (2)

$\sqrt{a}-\sqrt{b}=3$

$\sqrt{a}+\sqrt{b}=2$

Adding above two equations, we get,

$2 \sqrt{a}+0=2+3$

$\begin{array}{l}{2 \sqrt{a}=5} \\\\ {\sqrt{a}=\frac{5}{2}} \\\\ {a=\frac{25}{4}}\end{array}$

substitute $\sqrt{a}$ value in (2)

$\begin{array}{l}{\frac{5}{2}+\sqrt{b}=2} \\\\ {\sqrt{b}=\frac{2}{\sin \frac{5}{2}}} \\\\ {\sqrt{b}=\frac{4-5}{2}} \\\\ {\sqrt{b}=\frac{-1}{2}} \\\\ {b=\frac{1}{4}}\end{array}$