The most efficient method to solve a system quadratic-quadratic equations algebraically is elimination.
The following steps would be useful to solve the system by elimination.
Step 1 :
Add or subtract to eleiminate y.
(To eleiminate y, some times, you may have to multiply one of the equations or both equations by some constant depending on the coefficient of y in both the equations.)
Step 2 :
Solve the resulting equation for x.
Step 3 :
Substitute the x-values from step 2 into one of the two equations and solve for y.
Solve the systems of quadratic-quadratic equations algebraically and verify the solutions :
Example 1 :
2x^{2} - 16x - y = -35
2x^{2} - 8x - y = -11
Solution :
2x^{2} - 16x - y = -35 ----(1)
2x^{2} - 8x - y = -11 ----(2)
In the above system of two quadratic equations, y has the same coefficent and same sign.
By simply subtracting the two equations, y can be eliminated easily.
(2) - (1) :
(2x^{2} - 8x - y) - (2x^{2} - 16x - y) = -11 - (-35)
2x^{2} - 8x - y - 2x^{2} + 16x + y = -11 + 35
Group the like terms ang combine them.
(2x^{2 }- 2x^{2}) + (-8x + 16x) + (-y + y) = 24
0 + 8x + 0 = 24
8x = 24
Divide both sides by 8.
⁸ˣ⁄₈ = ²⁴⁄₈
x = 3
Substitute x = 3 into (1) and solve for y.
2(3)^{2} - 16(3) - y = -35
2(9) - 48 - y = -35
18 - 48 - y = -35
-30 - y = -35
Add 30 to both sides.
-y = -5
Multiply both sides by -1.
y = 5
The solution is (3, 5).
Verification :
Verify the solution (3, 5).
Substitu x = 3 and y = 5 into the original equations.
2x^{2} - 16x - y = -35 :
2(3)^{2} - 16(3) - 5 = -35
2(9) - 48 - 5 = -35
18 - 48 - 5 = -35
-35 = -35 ✔
2x^{2} - 8x - y = -11 :
2(3)^{2} - 8(3) - 5 = -11
2(9) - 24 - 5 = -11
18 - 24 - 5 = -11
-11 = -11 ✔
The solution (3, 5) is correct.
Example 2 :
2x^{2} + 16x + y = -26
x^{2} + 8x - y = -19
Solution :
2x^{2} + 16x + y = -26 ----(1)
x^{2} + 8x - y = -19 ----(2)
In the above system of two quadratic equations, y has the same coefficent with different signs.
By simply adding the two equations, y can be eliminated easily.
(1) + (2) :
(2x^{2} + 16x + y) + (x^{2} + 8x - y) = -26 + (-19)
2x^{2} + 16x + y + x^{2} + 8x - y = -26 - 19
Group the like terms ang combine them.
(2x^{2 }+ x^{2}) + (16x + 8x) + (y - y) = -45
3x^{2} + 24x + 0 = -45
3x^{2} + 24x = -45
Add 45 to both sides.
3x^{2} + 24x + 45 = 0
3(x^{2} + 8x + 15) = 0
Divide both sides by 3.
x^{2} + 8x + 15 = 0
Factor and solve it.
x^{2} + 5x + 3x + 15 = 0
x(x + 5) + 3(x + 5) = 0
(x + 5)(x + 3) = 0
x + 5 = 0 or x + 3 = 0
x = -5 or x = -3
Substitute x = -5 into (2). (-5)^{2} + 8(-5) - y = -19 25 - 40 - y = -19 -15 - y = -19 -y = -4 y = 4 |
Substitute x = -3 into (2). (-3)^{2} + 8(-3) - y = -19 9 - 24 - y = -19 -15 - y = -19 -y = -4 y = 4 |
The solutions are (-5, 4) and (-3, 4).
Verification :
Verify the solution (-5, 4).
Substitu x = -5 and y = 4 into the original equations.
2x^{2} + 16x + y = -26 :
2(-5)^{2} + 16(-5) + 4 = -26
2(25) - 80 + 4 = -26
50 - 80 + 4 = -26
-26 = -26 ✔
x^{2} + 8x - y = -19 :
(-5)^{2} + 8(-5) - 4 = -19
25 - 40 - 4 = -19
-19 = -19 ✔
Verify the solution (-3, 4).
Substitu x = -3 and y = 4 into the original equations.
2x^{2} + 16x + y = -26 :
2(-3)^{2} + 16(-3) + 4 = -26
2(9) - 48 + 4 = -26
18 - 48 + 4 = -26
-26 = -26 ✔
x^{2} + 8x - y = -19 :
(-3)^{2} + 8(-3) - 4 = -19
9 - 24 - 4 = -19
-19 = -19 ✔
Both solutions are correct.
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