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3 years ago

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Find the solution of this system of equations 10x-7y=-1 and -2x-7y=-25

1 answer:

Sergio039 [100]3 years ago

8 0

Answer:

<h2>x = 2, y = 3 → (2, 3)</h2>

Step-by-step explanation:

$\left\{\begin{array}{ccc}10x-7y=-1\\-2x-7y=-25&\text{change the signs}\end{array}\right\\\underline{+\left\{\begin{array}{ccc}10x-7y=-1\\2x+7y=25\end{array}\right}\qquad\text{add both sides of the equations}\\.\qquad12x=24\qquad\text{divide both sides by 12}\\.\qquad x=2\\\\\text{Put it to the second equation}\\2(2)+7y=25\\4+7y=25\qquad\text{subtract 4 from both sides}\\7y=21\qquad\text{divide both sides by 7}\\y=3$

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3 years ago

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3 years ago

-3x + 9 = -18 what's the answer

Art [367]

Answer:

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Step-by-step explanation:

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3 years ago

Read 2 more answers

Triangle ABC has AB=5, BC=7, and AC=9. D is on AC with BD=5. Find the length of DC

e-lub [12.9K]

Answer:

$DC=\frac{8}{3}\ units$

Step-by-step explanation:

The picture of the question in the attached figure

we know that

The triangle ABD is an isosceles triangle

because

AB=BD

The segment BM is a perpendicular bisector segment AD

so

<em>In the right triangle ABM</em>

Applying the Pythagorean Theorem

$BM^2=AB^2-AM^2$

we have

$AB=5\ units\\AM=x\ units$

substitute

$BM^2=5^2-x^2$

$BM^2=25-x^2$ -----> equation A

<em>In the right triangle BMC</em>

Applying the Pythagorean Theorem

$BM^2=BC^2-MC^2$

we have

$BC=7\ units\\MC=AC-AM=(9-x)\ units$

substitute

$BM^2=7^2-(9-x)^2$

$BM^2=49-(81-18x+x^2)$

$BM^2=49-81+18x-x^2$

$BM^2=-x^2+18x-32$ ----> equation B

equate equation A and equation B

$-x^2+18x-32=25-x^2$

solve for x

$18x=25+32\\18x=57\\\\x=\frac{57}{18}$

Simplify

$x=\frac{19}{6}$

<em>Find the length of DC</em>

$DC=AC-2x$

substitute the given values

$DC=9-2(\frac{19}{6})$

$DC=9-\frac{19}{3}\\\\DC=\frac{8}{3}\ units$

8 0

3 years ago