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klasskru [66]
3 years ago
15

Hey guys! I need help solving this equation! The equation is:

Mathematics
1 answer:
user100 [1]3 years ago
3 0

Ok so you have your two equations:
(1) 2x + 3y = -11 (I am assuming you meant 2x + 3y instead of 2x + 3x at least?)
(2) x - y = 12
x = y + 12

If we substitute equation (2) into equation (1) we get:
2(y + 12) + 3y = -11
2y + 24 + 3y = -11
5y = -35, therefor y = -7

If y = -7, from equation (2) we get:
x = -7 + 12 = 5

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denis-greek [22]
Do you need a domain and range? 
8 0
3 years ago
2x - 4y = 10
Bas_tet [7]

Answer:

Substitution, but this system of equations has no solution.

Step-by-step explanation:

2x-4y=10

2y+6=x

----------------

2(2y+6)-4y=10

4y+12-4y=10

12=10

no solution.

5 0
3 years ago
What are some examples of Adjacent Congruent Angles.
Anastasy [175]

Answer:

  • base angles of an isosceles triangle
  • angles at successive corners of any regular polygon
  • angles either side of an angle bisector
  • angles made by a perpendicular to a line (excluding vertical angles)

Step-by-step explanation:

When it comes to angles, the term "adjacent" means different things in different contexts. Any pair of angles in a triangle are considered to be adjacent. The base angles of an isosceles triangle are adjacent congruent angles.

Angles at successive corners of any regular polygon are congruent adjacent angles.

Angle that have a common vertex and a common side are also called adjacent angles. Angles on either side of an angle bisector are adjacent congruent angles. Likewise any non-vertical pair of angles where lines cross at right angles are adjacent congruent angles.

3 0
3 years ago
In parallelogram DEFG, DH = x + 1, HF = 3y, G H = 3 x − 4 , and HE = 5y + 1. Find the values of x and y. The diagram is not draw
Nana76 [90]

Answer:

see explanation

Step-by-step explanation:

Using the property of parallelograms

• The diagonals bisect each other, hence

DH = HF and GH = HE

x + 1 = 3y and 3x - 4 = 5y + 1 ⇒ 3x = 5y + 5

Solving the 2 equations simultaneously

x + 1 = 3y → (1)

3x = 5y + 5 → (2)

rearrange (1) expressing x in terms of y

x = 3y - 1 → (3)

substitute x = 3y - 1 in (2)

3(3y - 1) = 5y + 5

9y - 3 = 5y + 5 ( subtract 5y from both sides )

4y - 3 = 5 ( add 3 to both sides )

4y = 8 ( divide both sides by 4 )

y = 2

substitute y = 2 into (3)

x = (3 × 2) - 1 = 6 - 1 = 5

Hence x = 5, y = 2



3 0
3 years ago
16. Interest earned:
photoshop1234 [79]

\bf ~~~~~~ \textit{Simple Interest Earned} \\\\ I = Prt\qquad \begin{cases} I=\textit{interest earned}\\ P=\textit{original amount deposited}\dotfill & \$1250\\ r=rate\to 3\%\to \frac{3}{100}\dotfill &0.03\\ t=years\dotfill &4 \end{cases} \\\\\\ I=(1250)(0.03)(4)\implies I=150

3 0
3 years ago
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