Download Use the substitution method to find all solutions of the

Use the substitution method to find all solutions of the system equations.
1. 2x + y = 7
x + 2y = 2
Solve first equation for y:
2x – 2x + y = 7 – 2x
y = 7 – 2x
Substitute this into the second equation:
x + 2(7 – 2x) = 2
x + 14 – 4x = 2
x – 4x = 2 – 14
-3x = -12
x=4
Substitute this in the equation for y:
y = 7 – 2(4)
y=7–8
y = -1
The solution is: (4, -1)
2. x2 + y2 = 25
y = 2x
Using the second equation, substitute for y in the first equation:
x2 + (2x)2 = 25
x2 + 4x2 = 25
5x2 = 25
x2 = 5
x = 5, " 5
Now, using the second equation:
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x= 5:
y = 2x = 2 5 = 2 5
x=" 5:
y = 2x = 2 " 5 = "2 5
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There solutions are:
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5, 2 5 , " 5, " 2 5
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3. x2 + y = 9
x–y+3=0
Solve the second equation for x:
x–y+3=0
x–y+y+3=0+y
x+3=y
x+3–3=y–3
x=y-3
Substitute this into the first equation:
(y – 3)2 + y = 9
y2 – 6y + 9 + y = 9
y2 – 5y + 9 = 9
y2 – 5y + 9 – 9 = 9 – 9
y2 – 5y = 0
y(y – 5) = 0
y = 0, and y = 5
Substitute these in the above equation for x:
y = 0: x = y – 3 = 0 – 3 = -3
y = 5: x = y – 3 = 5 – 3 = 2
The solutions are (-3, 0) and (2, 5)
4. x2 – y = 1
2x2 + 3y = 17
Solve the first equation for y:
x2 – y + y = 1 + y
x2 = 1 + y
x2 – 1 = 1 – 1 + y
y = x2 – 1
Substitute this in the second equation:
2x2 + 3(x2 – 1) = 17
2x2 + 3x2 – 3 = 17
5x2 – 3 = 17
5x2 – 3 + 3 = 17 + 3
5x2 = 20
x2 = 4
x = 2, -2
Substituting these in the equation for y:
x = 2:
y = x2 – 1 = 22 – 1 = 4 – 1 = 3
x = -2:
y = x2 – 1 = (-2)2 – 1 = 4 – 1 = 3
The solutions are (2, 3) and (-2, 3)
use the elimination method to find all solutions of the system of equations.
1. 4x – 3y = 11
8x + 4y = 12
Multiply the first equation by 2 to get:
2(4x – 3y) = 2(11)
8x – 6y = 22
Subtract this equation from the second given equation to get:
(8x – 8x) + (4y – (-6y)) = 12 – 22
10y = -10
y = -10/10
y = -1
Substitute this result into the first equation and solve for x:
4x – 3(-1) = 11
4x + 3 = 11
4x = 11 – 3
4x = 8
x=2
The solution is (2, -1)
2. 3x2 + 4y = 17
2x2 + 5y = 2
Multiply the first equation by 2, and the second equation by 3 to get:
2(3x2 + 4y) = 2(17)
3(2x2 + 5y) = 3(2)
6x2 + 8y = 34
6x2 + 15y = 6
Subtract the second equation from the first:
6x2 – 6x2 + 8y – 15y = 34 – 6
-7y = 28
y = -4
Substitute this into the first equation and solve for x:
3x2 + 4(-4) = 17
3x2 – 16 = 17
3x2 – 16 + 16 = 17 + 16
3x2 = 33
x2 = 11
x = 11, " 11
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The solutions are
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11, " 4 , " 11, " 4
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3. 2x2 + 4y = 13
x2 – y2 = 7/2
Multiply the second equation by 2 to get the following system:
2x2 + 4y = 13
2(x2 – y2) = 2(7/2)
2x2 + 4y = 13
2x2 – 2y2 = 7
Subtract the second equation from the first to get:
2x2 – 2x2 + 4y – (-2y2) = 13 – 7
4y + 2y2 = 6
2y2 + 4y – 6 = 0
This can be factored into:
2(y2 + 2y – 3) = 0
2(y – 1)(y + 3) = 0
Setting each factor equal to zero, using the Principle of Zero Products:
y–1=0
y=1
and
y+3=0
y = -3
Substituting each of these solutions for y into the first of the given equation gives:
y = 1:
2x2 + 4(1) = 13
2x2 + 4 = 13
2x2 = 9
x2 = 9/2
x=
9
3
3 2
=±
=±
2
2
2
y = -3
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2x2 + 4(3) = 13
2x2 + 12 = 13
2x2 = 1
x2 = 1/2
x=
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The four solutions
are then:
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1
2
=±
=±
2
2
2
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4. x2 – y2 = 1
2x2 – y2 = x + 3
Subtract the first equation from the second equation to get:
2x2 – x2 – y2 – (-y2) = x + 3 – 1
x2 = x + 2
x2 – x – 2 = 0
This can be factored as:
(x + 1)(x – 2) = 0
Setting each factor equal to zero, using the Principle of Zero Products:
x +1=0
x = -1
and
x–2=0
x=2
Substituting these values for x into the first given equation and solving for y
gives:
(-1)2 – y2 = 1
1 – y2 = 1
y2 = 0
y=0
This solution is (-1, 0)
(2)2 – y2 = 1
4 – y2 = 1
y2 = 4 – 1
y2 = 3
y = 3, " 3
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3 and 2, " 3
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3 and 2, " 3
The solutions are 2,
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The solutions are (1, 0), 2,
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find the first five terms and the 100th term of the sequence.
1. an = 2n + 3
n = 1:
n = 2:
n = 3:
n = 4:
n = 5:
n = 100:
a1 = 2(1) + 3 = 2 + 3 = 5
a2 = 2(2) + 3 = 4 + 3 = 7
a3 = 2(3) + 3 = 6 + 3 = 9
a4 = 2(4) + 3 = 8 + 3 = 11
a5 = 2(5) + 3 = 10 + 3 = 13
a100 = 2(100) + 3 = 200 + 3 = 203
2. an = n2 + 1
n = 1:
n = 2:
n = 3:
n = 4:
n = 5:
n = 100:
a1 = (1)2 + 1 = 1 + 1 = 2
a2 = (2)2 + 1 = 4 + 1 = 5
a3 = (3)2 + 1 = 9 + 1 = 10
a4 = (4)2 + 1 = 16 + 1 = 17
a5 = (5)2 + 1 = 25 + 1 = 26
a100 = (100)2 + 1 = 10,000 + 1 = 10,001
3. an = 1/n2
n = 1:
n = 2:
n = 3:
n = 4:
n = 5:
n = 100:
a1 = 1/(1)2 = 1/1 = 1
a2 = 1/(2)2 = 1/4
a3 = 1/(3)2 = 1/9
a4 = 1/(4)2 = 1/16
a5 = 1/(5)2 = 1/25
a100= 1/(100)2 = 1/10,000
4. an = (-1)^n+1 [n/n+1]
n = 1:
n = 2:
n = 3:
n = 4:
n = 5:
n = 100:
a1 = (-1)1+1[1/(1+1)] = (-1)2[1/2] = 1/2
a2 = (-1)2+1[1/(2+1)] = (-1)3[1/3] = -1/3
a3 = (-1)3+1[1/(3+1)] = (-1)4[1/4] = 1/4
a4 = (-1)4+1[1/(4+1)] = (-1)5[1/5] = -1/5
a5 = (-1)5+1[1/(5+1)] = (-1)6[1/6] = 1/6
a100= (-1)100+1[1/(100+1)] = (-1)101[1/101] = -1/101
5. an = 3
n = 1:
n = 2:
n = 3:
a1 = 3
a2 = 3
a3 = 3
n = 4:
n = 5:
n = 100:
a4 = 3
a5 = 3
a100 = 3