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Rainbow [258]
3 years ago
13

Solve by substitution 3x+y=3 and 7x+2y=1

Mathematics
1 answer:
balu736 [363]3 years ago
5 0

3x + y = 3

7x + 2y = 1

First isolate one of the variables (x or y) in one of the equations.

Isolate "y" in the first equation(because it is the easiest to isolate) and substitute it into the second equation.

3x + y = 3      Subtract 3x on both sides

3x - 3x + y = 3 - 3x

y = 3 - 3x


7x + 2y = 1

7x + 2(3 - 3x) = 1     [since y = 3 - 3x, you can substitute (3-3x) for "y"]

Multiply/distribute 2 into (3 - 3x)

7x + (3(2) - 3x(2)) = 1

7x + 6 - 6x = 1

x + 6 = 1     Subtract 6 on both sides

x = -5


Now that you know "x", substitute it into one of the equations (I will do both)

3x + y = 3

3(-5) + y = 3       [since x = -5, you can plug in -5 for "x"]

-15 + y = 3    Add 15 on both sides

y = 18


7x + 2y = 1

7(-5) + 2y = 1

-35 + 2y = 1     Add 35 on both sides

2y = 36     Divide 2 on both sides

y = 18


x = -5, y = 18  or  (-5, 18)

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Find the equation of the median from b in ABC whose vertices are (1,5), B(5,3) and C(-3, -2)
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Answer:

y = x + 6

x = 1

y = ¼(x - 5) + 3

Step-by-step explanation:

Vetices are;

A(1,5), B(5,3) and C(-3, -2)

Thus;

Median of AB is; D = (1 + 5)/2, (5 + 3)/2

D = (3, 4)

Median of BC is; E = (5 + (-3))/2, (3 + (-2))/2

E = (1, 0.5)

Median of AC is; F ; (-3 + 1)/2, (-2 + 5)/2

F = (-1, 1.5)

Thus, the median lines will be;

CD, AE & BF.

Thus;

Equation of CD is;

(y - (-3))/(x - (-2)) = (-2 - 4))/(-3 - 3)

(y + 4)/(x + 2) = -6/-6

y - 4 = 1(x + 2)

y = 4 + x + 2

y = x + 6

Equation of AE;

(y - 5)/(x - 1) = (0.5 - 5)/(1 - 1)

(y - 5)/(x - 1) = -4.5/0

Cross multiply to get;

0(y - 5) = -4.5(x - 1)

-4.5x = -4.5

x = 1

Equation of BF;

(y - 3)/(x - 5) = (1.5 - 3)/(-1 - 5)

(y - 3)/(x - 5) = -1.5/-6

(y - 3)/(x - 5) = 1/4

y - 3 = ¼(x - 5)

y = ¼(x - 5) + 3

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