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3 years ago

Use elimination to solve each system of equations 2x-y=-1 3x-2y=1

2 answers:

8 0

$2x-y=-1 \ \ \ |\times (-2) \\
3x-2y=1 \\ \\
-4x+2y=2 \\
\underline{3x-2y=1 \ \ \ \ } \\
-4x+3x=2+1 \\
-x=3 \ \ \ |\times (-1) \\
x=-3 \\ \\
2x-y=-1 \\
2 \times (-3)-y=-1 \\
-6-y=-1 \ \ \ |+6 \\
-y=5 \ \ \ |\times (-1) \\
y=-5 \\ \\
\boxed{(x,y)=(-3,-5)}$

TEA [102]3 years ago

4 0

2x - y = -1
3x - 2y = 1
5x - 3y = 0
5x - 5x - 3y = 0 - 5x
3y = 0 - 5x
3y = 0 - 5x
3 3
y = 0 - 5/3x
3x - 2(0 - 5/3x) = 1
3x + 8/3x = 1
5 1/4x = 1
5 1/4x = 1 
5 1/4 5 1/4
 x = 4/21
3(4/21) - 2y = 1
4/7 - 2y = 1
-4/7 -4/7
 -2y = 3/7
 -2y = 3/7
 -2 -2
 y = -3/14
(x, y) = (4/21, -3/14)

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Find the area of a quadrilateral ABCD in which AB = 3 cm, BC = 4 cm, CD = 4 cm, DA = 5 cm and AC = 5 cm.

melamori03 [73]

Answer:

$6+2\sqrt{21}\:\mathrm{cm^2}\approx 15.17\:\mathrm{cm^2}$

Step-by-step explanation:

The quadrilateral ABCD consists of two triangles. By adding the area of the two triangles, we get the area of the entire quadrilateral.

Vertices A, B, and C form a right triangle with legs $AB=3$ , $BC=4$ , and $AC=5$ . The two legs, 3 and 4, represent the triangle's height and base, respectively.

The area of a triangle with base $b$ and height $h$ is given by $A=\frac{1}{2}bh$ . Therefore, the area of this right triangle is:

$A=\frac{1}{2}\cdot 3\cdot 4=\frac{1}{2}\cdot 12=6\:\mathrm{cm^2}$

The other triangle is a bit trickier. Triangle $\triangle ADC$ is an isosceles triangles with sides 5, 5, and 4. To find its area, we can use Heron's Formula, given by: