
Given x = 2y + 5 and y = (2x - 3)(x + 9).
How many ordered pairs (x, y) satisfy the system of equations shown above?
A. 0
B. 1
C. 2
D. Infinitely many
Answer
466.8k+ views
Hint: A polynomial of degree n, has at most n roots.
Not all solutions may satisfy the given conditions, and some solutions may be repeated.
In order to solve a system of equations with two variables, we find the expression for one of the variables in terms of the other by using one of the equations and then substitute it in the other equation.
The two solutions of the general quadratic equation $a{{x}^{2}}+bx+c=0$ , are given by:
$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
(i)If ${{b}^{2}}-4ac>0$ , then there are two distinct solutions.
(ii)If ${{b}^{2}}-4ac=0$ , then there are two solutions, each equal to $\dfrac{-b}{2a}$ .
Complete step-by-step answer:
Let us form an equation involving only one of the variables, either x or y.
It is given that x = 2y + 5.
Substituting this value of x in the equation y = (2x - 3)(x + 9), we will get:
⇒ y = [2(2y + 5) - 3][(2y + 5) + 9}]
On removing the inside brackets "()" by performing multiplication, we will get:
⇒ y = [4y + 10 - 3][2y + 5 + 9]
⇒ y = (4y + 7)(2y + 14)
On expanding by multiplying the two terms using the distributive property of multiplication:
⇒ $y=8{{y}^{2}}+56y+14y+98$
⇒ $8{{y}^{2}}+69y+98=0$
Comparing this equation with the general equation $a{{x}^{2}}+bx+c=0$ , we can say that a = 8, b = 69 and c = 98.
Let us calculate the value of ${{b}^{2}}-4ac$ by substituting the values of a, b and c:
${{b}^{2}}-4ac={{69}^{2}}-4(8)(98)=4761-3136=1625$
From 1625 > 0, we will get two distinct values of y, say ${{y}_{1}}$ and ${{y}_{2}}$ .
And, since the degree of x in x = 2y + 5 is one, we will get only one value of x for each value of y, say ${{x}_{1}}$ for ${{y}_{1}}$ and ${{x}_{2}}$ for ${{y}_{2}}$ .
The order pairs satisfying the above equations are therefore, $({{x}_{1}},{{y}_{1}})$ and $({{x}_{2}},{{y}_{2}})$ .
Therefore, the number of ordered pairs which satisfy the given system of equations is two.
Note: "How many values of x?" and "What is the value of x?" are two significantly different questions.
We do not need to calculate the exact values of x and y for answering this question, because it asks "How many (x, y)?" only.
The degree of a polynomial/equation is the highest power of the variables occurring in it.
(xy) has a degree of two, because the variables x and y are multiplied together.
An ordered pair is different from other types of pairs in the sense that the order in which the objects appear in the pair is significant.
So, (a, b) and (b, a) are two different ordered pairs.
Not all solutions may satisfy the given conditions, and some solutions may be repeated.
In order to solve a system of equations with two variables, we find the expression for one of the variables in terms of the other by using one of the equations and then substitute it in the other equation.
The two solutions of the general quadratic equation $a{{x}^{2}}+bx+c=0$ , are given by:
$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
(i)If ${{b}^{2}}-4ac>0$ , then there are two distinct solutions.
(ii)If ${{b}^{2}}-4ac=0$ , then there are two solutions, each equal to $\dfrac{-b}{2a}$ .
Complete step-by-step answer:
Let us form an equation involving only one of the variables, either x or y.
It is given that x = 2y + 5.
Substituting this value of x in the equation y = (2x - 3)(x + 9), we will get:
⇒ y = [2(2y + 5) - 3][(2y + 5) + 9}]
On removing the inside brackets "()" by performing multiplication, we will get:
⇒ y = [4y + 10 - 3][2y + 5 + 9]
⇒ y = (4y + 7)(2y + 14)
On expanding by multiplying the two terms using the distributive property of multiplication:
⇒ $y=8{{y}^{2}}+56y+14y+98$
⇒ $8{{y}^{2}}+69y+98=0$
Comparing this equation with the general equation $a{{x}^{2}}+bx+c=0$ , we can say that a = 8, b = 69 and c = 98.
Let us calculate the value of ${{b}^{2}}-4ac$ by substituting the values of a, b and c:
${{b}^{2}}-4ac={{69}^{2}}-4(8)(98)=4761-3136=1625$
From 1625 > 0, we will get two distinct values of y, say ${{y}_{1}}$ and ${{y}_{2}}$ .
And, since the degree of x in x = 2y + 5 is one, we will get only one value of x for each value of y, say ${{x}_{1}}$ for ${{y}_{1}}$ and ${{x}_{2}}$ for ${{y}_{2}}$ .
The order pairs satisfying the above equations are therefore, $({{x}_{1}},{{y}_{1}})$ and $({{x}_{2}},{{y}_{2}})$ .
Therefore, the number of ordered pairs which satisfy the given system of equations is two.
Note: "How many values of x?" and "What is the value of x?" are two significantly different questions.
We do not need to calculate the exact values of x and y for answering this question, because it asks "How many (x, y)?" only.
The degree of a polynomial/equation is the highest power of the variables occurring in it.
(xy) has a degree of two, because the variables x and y are multiplied together.
An ordered pair is different from other types of pairs in the sense that the order in which the objects appear in the pair is significant.
So, (a, b) and (b, a) are two different ordered pairs.
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