Problems 1-2 : Solve the given system of linear and quadratic equations by substitution and verify your solution.
Problems 1 :
5x - y - 10 = 0
x2 + x - 2y = 0
Problems 2 :
3x + y + 9 = 0
4x2 - x + y + 9 = 0
Problems 3-4 : Solve the given system of linear and quadratic equations by elimination and verify your solution.
Problems 3 :
x - y + 6 = 0
x2 + 3x + y - 2 = 0
Problems 4 :
2x + y - 2 = 0
x2 - 2x + y - 7 = 0
1. Answer :
5x - y - 10 = 0 ----(1)
x2 + x - 2y = 0 ----(2)
In (1), solve for y in terms of x.
5x - y - 10 = 0
-y = -5x + 10
y = 5x - 10 ----(3)
Substitute '5x - 10' for y in (2).
x2 + x - 2(5x - 10) = 0
x2 + x - 10x + 20 = 0
x2 - 9x + 20 = 0
Factor and solve.
x2 - 4x - 5x + 20 = 0
x(x - 4) - 5(x - 4) = 0
(x - 4)(x - 5) = 0
x - 4 = 0 or x - 5 = 0
x = 4 or x = 5
Substitute x = 4 in (3). y = 5(4) - 10 y = 20 - 10 y = 10 |
Substitute x = 5 in (3). y = 5(5) - 10 y = 25 - 10 y = 15 |
The two solutions are (4, 10) and (5, 15).
Verification :
Verify the solution (4, 10).
Substitu x = 4 and y = 10 into the original equations.
5x - y - 10 = 0 :
5(4) - 10 - 10 = 0
20 - 10 - 10 = 0
0 = 0 ✔
x2 + x - 2y = 0 :
42 + 4 - 2(10) = 0
16 + 4 - 20 = 0
0 = 0 ✔
Verify the solution (5, 15).
Substitu x = 5 and y = 15 into the original equations.
5x - y - 10 = 0 :
5(5) - 15 - 10 = 0
25 - 15 - 10 = 0
0 = 0 ✔
x2 + x - 2y = 0 :
52 + 5 - 2(15) = 0
25 + 5 - 30 = 0
0 = 0 ✔
Both solutions are correct.
2. Answer :
3x + y + 9 = 0 ----(1)
4x2 - x + y + 9 = 0 ----(2)
In (1), solve for y in terms of x.
3x + y + 9 = 0
y = -3x - 9 ----(3)
Substitute '-3x - 9' for y in (2).
4x2 - x + (-3x - 9) + 9 = 0
4x2 - x - 3x - 9 + 9 = 0
4x2 - 4x = 0
Factor and solve.
4x(x - 1) = 0
4x = 0 or x - 1 = 0
x = 0 or x = 1
Substitute x = 0 in (3). y = -3(0) - 9 y = 0 - 9 y = -9 |
Substitute x = 1 in (3). y = -3(1) - 9 y = -3 - 9 y = -12 |
The two solutions are (0, -9) and (1, -12).
Verification :
Verify the solution (0, -9).
Substitu x = 0 and y = -9 into the original equations.
3x + y + 9 = 0 :
3(0) + (-9) + 9 = 0
0 - 9 + 9 = 0
0 = 0 ✔
4x2 - x + y + 9 = 0 :
4(0)2 - 0 + (-9) + 9 = 0
0 + 0 - 9 + 9 = 0
0 = 0 ✔
Verify the solution (1, -12).
Substitu x = 1 and y = -12 into the original equations.
3x + y + 9 = 0 :
3(1) + (-12) + 9 = 0
3 - 12 + 9 = 0
0 = 0 ✔
4x2 - x + y + 9 = 0 :
4(1)2 - 1 + (-12) + 9 = 0
4 - 1 - 12 + 9 = 0
0 = 0 ✔
Both solutions are correct.
3. Answer :
x - y + 6 = 0 ----(1)
x2 + 3x + y - 2 = 0 ----(2)
By adding the two equations above, y can be eliminated.
(1) + (2) :
(x - y + 6) + (x2 + 3x + y - 2) = 0
x - y + 6 + x2 + 3x + y - 2 = 0
x2 + 4x + 4 = 0
Factor and solve.
x2 + 2x + 2x + 4 = 0
x(x + 2) + 2(x + 2) = 0
(x + 2)(x + 2) = 0
(x + 2)2 = 0
Taking square root on both sides,
x + 2 = 0
x = -2
Substitute x = -2 in (1).
-2 - y + 6 = 0
-y + 4 = 0
-y = -4
y = 4
The solution is (-2, 4).
Verification :
Verify the solution (-2, 4).
Substitu x = -2 and y = 4 into the original equations.
x - y + 6 = 0
x - y + 6 = 0
-2 - 4 + 6 = 0
0 = 0 ✔
x2 + 3x + y - 2 = 0
x2 + 3x + y - 2 = 0
(-2)2 + 3(-2) + 4 - 2 = 0
4 - 6 + 4 - 2 = 0
0 = 0 ✔
The solution is (-2, 4) is correct.
4. Answer :
2x + y - 2 = 0 ----(1)
x2 - 2x + y - 7 = 0 ----(2)
By subtracting the two equations above, y can be eliminated.
(2) - (1) :
(x2 - 2x + y - 7) - (2x + y - 2) = 0
x2 - 2x + y - 7 - 2x - y + 2 = 0
x2 - 4x - 5 = 0
Factor and solve.
x2 - 5x + x - 5 = 0
x(x - 5) + 1(x - 5) = 0
(x - 5)(x + 1) = 0
x - 5 = 0 or x + 1 = 0
x = 5 or x = -1
Substitute x = 5 in (1). 2(5) + y - 2 = 0 10 + y - 2 = 0 y + 8 = 0 y = -8 |
Substitute x = -1 in (1). 2(-1) + y - 2 = 0 -2 + y - 2 = 0 y - 4 = 0 y = 4 |
The two solutions are (5, -8) and (-1, 4).
Verification :
Verify the solution (5, -8).
Substitu x = 5 and y = -8 into the original equations.
2x + y - 2 = 0
2(5) + (-8) - 2 = 0
10 - 8 - 2 = 0
0 = 0 ✔
x2 - 2x + y - 7 = 0
x2 - 2x + y - 7 = 0
52 - 2(5) + (-8) - 7 = 0
25 - 10 - 8 - 7 = 0
0 = 0 ✔
Verify the solution (-1, 4).
Substitu x = -1 and y = 4 into the original equations.
2x + y - 2 = 0
2(-1) + 4 - 2 = 0
-2 + 4 - 2 = 0
0 = 0 ✔
x2 - 2x + y - 7 = 0
(-1)2 - 2(-1) + 4 - 7 = 0
1 + 2 + 4 - 7 = 0
0 = 0 ✔
Both solutions are correct.
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