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Artyom0805 [142]
3 years ago
8

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

Mathematics
2 answers:
mixer [17]3 years ago
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
<u>3x - 2y = 1</u>
5x - 3y = 0
5x - 5x - 3y = 0 - 5x
3y = 0 - 5x
<u>3y</u> = <u>0 - 5x</u>
 3        3
y = 0 - 5/3x
3x - 2(0 - 5/3x) = 1
3x + 8/3x = 1
5 1/4x = 1
<u>5 1/4x </u>=<u>  1    </u>
5 1/4    5 1/4
     x = 4/21
3(4/21) - 2y = 1
 4/7 - 2y = 1
<u>-4/7         -4/7</u>
         -2y = 3/7
         -<u>2y</u> = <u>3/7</u>
          -2      -2
           y = -3/14
(x, y) = (4/21, -3/14)
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Tomika heard that the diagonals of a rhombus are perpendicular to each other. Help her test her conjecture. Graph quadrilateral
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Answer:

a. The four sides of the quadrilateral ABCD are equal, therefore, ABCD is a rhombus

b. The equation of the diagonal line AC is y = 5 - x

The equation of the diagonal line BD is y = 5 - x

c. The diagonal lines AC and BD of the quadrilateral ABCD are perpendicular to each other

Step-by-step explanation:

The vertices of the given quadrilateral are;

A(1, 4), B(6, 6), C(4, 1) and D(-1, -1)

a. The length, l, of the sides of the given quadrilateral are given as follows;

l = \sqrt{\left (y_{2}-y_{1}  \right )^{2}+\left (x_{2}-x_{1}  \right )^{2}}

The length of side AB, with A = (1, 4) and B = (6, 6) gives;

l_{AB} = \sqrt{\left (6-4  \right )^{2}+\left (6-1  \right )^{2}} = \sqrt{29}

The length of side BC, with B = (6, 6) and C = (4, 1) gives;

l_{BC} = \sqrt{\left (1-6  \right )^{2}+\left (4-6  \right )^{2}} = \sqrt{29}

The length of side CD, with C = (4, 1) and D = (-1, -1) gives;

l_{CD} = \sqrt{\left (-1-1  \right )^{2}+\left (-1-4  \right )^{2}} = \sqrt{29}

The length of side DA, with D = (-1, -1) and A = (1,4)   gives;

l_{DA} = \sqrt{\left (4-(-1)  \right )^{2}+\left (1-(-1)  \right )^{2}} = \sqrt{29}

Therefore, each of the lengths of the sides of the quadrilateral ABCD are equal to √(29), and the quadrilateral ABCD is a rhombus

b. The diagonals are AC and BD

The slope, m, of AC is given by the formula for the slope of a straight line as follows;

Slope, \, m =\dfrac{y_{2}-y_{1}}{x_{2}-x_{1}}

Therefore;

Slope, \, m_{AC} =\dfrac{1-4}{4-1} = -1

The equation of the diagonal AC in point and slope form is given as follows;

y - 4 = -1×(x - 1)

y = -x + 1 + 4

The equation of the diagonal AC is y = 5 - x

Slope, \, m_{BD} =\dfrac{-1-6}{-1-6} = 1

The equation of the diagonal BD in point and slope form is given as follows;

y - 6 = 1×(x - 6)

y = x - 6 + 6 = x

The equation of the diagonal BD is y = x

c. Comparing the lines AC and BD with equations, y = 5 - x and y = x, which are straight line equations of the form y = m·x + c, where m = the slope and c = the x intercept, we have;

The slope m for the diagonal AC = -1 and the slope m for the diagonal BD = 1, therefore, the slopes are opposite signs

The point of intersection of the two diagonals is given as follows;

5 - x = x

∴ x = 5/2 = 2.5

y = x = 2.5

The lines intersect at (2.5, 2.5), given that the slopes, m₁ = -1 and m₂ = 1 of the diagonals lines satisfy the condition for perpendicular lines m₁ = -1/m₂, therefore, the diagonals are perpendicular.

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