Special Equations
If the number of variables in a system of equations is greater than the number of equations, solutions are typically sought under special conditions and restrictions.
Example:
\[\begin{aligned} a +2b + 3c = 14 \\ 4a + 3b + 2c = 16 \end{aligned}\]
Let’s find the value of the sum \( a + b + c \) in this system of equations.
If we add the equations side by side:
\[\begin{aligned} a +2b + 3c &= 14 \\ + \quad 4a + 3b + 2c &= 16 \\ \hline \\ 5a + 5b + 5c & = 30 \end{aligned}\]
\[ 5(a+b +c) = 30 \Rightarrow a+ b+ c = 6 \] is found.
Question 22
\[
\begin{aligned}
a- b &=22 \\
b+ c &= 10 \\
c-d&= 8
\end{aligned}
\]
then what is the expression \( a-2b-2c+ d \) equal to?
\[
\text{A)} 1 \quad
\text{B) } 2 \quad
\text{C) } 3 \quad
\text{D) } 4 \quad
\text{E) } 5
\]
Solution:
If both sides of the second and third equations are multiplied by \(-1 \) and the three equations are added side by side:
\[
\begin{aligned}
a- b &=22 \\
-b- c &=- 10 \\
+ \quad -c+ d&= -8\\
\hline
a-2b-2c+d &= 4
\end{aligned}
\]
\(\textbf{Answer: D} \)
Question 23
\[
\begin{aligned}
2a+ b + c =&12 \\
a- b + c= &4 \\
\end{aligned}
\]
then what is the expression \( \large \frac{3a+ 2c }{a+ 2b} \) equal to?
\[
\text{A)} 1 \quad
\text{B) } 2 \quad
\text{C) } 3 \quad
\text{D) } 4 \quad
\text{E) } 5
\]
Solution:
If the equations are added side by side and also subtracted side by side:
\[
\begin{aligned}
2a+ b + c =&12 \\
+\quad a- b + c= &4 \\ \hline \\ 3a+2c = &16
\end{aligned}
\]
\[
\begin{aligned}
2a+ b + c =&12 \\
-\quad a- b + c= &4 \\ \hline \\ a+2b = &8
\end{aligned}
\]
are found. Accordingly,
\[ \frac{3a+2c}{a+2b} = \frac{16}{8} = 2 \]
\(\textbf{Answer: B} \)
Question 24
\[
\begin{aligned}
a+ b =&9 \\
b + c= &8 \\ a^2-c^2 = &7
\end{aligned}
\]
then what is the sum \( a+ b+ c \) equal to?
\[
\text{A)} 8 \quad
\text{B) } 9 \quad
\text{C) } 10 \quad
\text{D) } 11 \quad
\text{E) } 12
\]
Solution:
\[ a^2-c^2 = 7 \Rightarrow (a-c ) \cdot (a+c) = 7 \]
\[
\begin{aligned}
a+ b =&9 \\
– \quad b + c= &8 \\ \hline \\ a-c = 1
\end{aligned}
\]
\[ (a-c ) \cdot (a+c) = 7 \Rightarrow 1 \cdot (a+ c ) = 7 \Rightarrow a + c = 7 \] is found. Therefore,
\[
\begin{aligned}
a+ b =&9 \\
b + c= &8 \\ \ + \quad a+ c = & 7 \\ \hline \\ 2a+ 2b+ 2c= &24
\end{aligned}
\]
\[ 2a+ 2b+ 2c= 24 \Rightarrow 2(a+b+c) = 24 \]
\[ \Rightarrow a+ b+c = 12\]
\(\textbf{Answer: E} \)
Question 25
\[
\begin{aligned}
2x-3y+ z =&-1 \\
x-2y+3z =& \;\;\;\; 6 \\
2y-x-2z = &-3
\end{aligned}
\]
then what is the difference \( x-y \) equal to?
\[
\text{A)} -1 \quad
\text{B) } 0 \quad
\text{C) } 1 \quad
\text{D) } 2 \quad
\text{E) } 3
\]
Solution:
After multiplying both sides of the third equation by 2, if the equations are added side by side:
\[
\begin{aligned}
2x-3y+ z =&-1 \\
x-2y+3z = &\;\; \;\;6 \\
+ \quad 4y-2x-4z =& -6\\
\hline\\
x-y= -1
\end{aligned}
\]
\(\textbf{Answer: A} \)
Question 26
\[
\begin{aligned}
\frac{a}{bc} = &\;2 \\
\frac{b}{ac} =& \;6 \\
\frac{c}{ab} = &\;5
\end{aligned}
\]
then what is the value of the product \( a \cdot b \cdot c \) equal to?
\[
\text{A)} \frac{1}{25} \quad
\text{B) } 25 \quad
\text{C) } \frac{1}{30} \quad
\text{D) } 30 \quad
\text{E) } \frac{1}{40}
\]
Solution:
Let’s multiply the three equations side by side.
\[ \frac{a}{bc} \cdot \frac{b}{ac} \cdot \frac{c}{ab} = 2 \cdot 3 \cdot 5 \]
\[ \frac{a\;b\;c}{a^2\; b^2\; c^2 } =30 \]
\[ \Rightarrow abc = \frac{1}{30} \]
\(\textbf{Answer: C} \)
Question 27
The equation \( (x-y-5)a + (x+y-3)b=0 \) is satisfied for every real value of a and b. Accordingly, what is the value of x?
\[
\text{A)} 1 \quad
\text{B) } 2 \quad
\text{C) } 3 \quad
\text{D) } 4 \quad
\text{E) } 5
\]
Solution:
If the given equation is satisfied for every real value of \(a \) and \(b \), then the expressions multiplied by \(a \) and \(b \) must be equal to zero.
\[
\begin{aligned}
x-y-5= &\;0 \\
+ \quad x+y-3 =& \;\;0 \\
\hline \\
2x-8=&0
\end{aligned}
\]
\[ 2x-8=0 \Rightarrow x= 4\]
\(\textbf{Answer: D} \)
Question 28
If \[ (x+y-2)^2 + (4x^2-4x+1)^3=0 \] then what is the value of \(y\)?
\[
\text{A)} 4 \quad
\text{B) } 3 \quad
\text{C) } \frac{5}{2} \quad
\text{D) } 2 \quad
\text{E) } \frac{3}{2}
\]
Solution:
\[ (x+y-2)^2 + (4x^2-4x+1)^3=0 \]
\[ \Rightarrow (x + y – 2)^2 + \left[(2x – 1)^2\right]^3 = 0 \]
\[ \Rightarrow (x+y-2)^2 + (2x-1)^6=0 \]
In this equation, since
\[ \Rightarrow (x + y – 2)^2 \ge 0 \quad \text{and} \quad (2x – 1)^6 \ge 0 \]
for the sum of these two expressions to be \(0\), we must have
\[(x + y – 2)^2 =0 \quad \text{and} \quad (2x – 1)^6 = 0 .\] From here,
\[ x+y-2= 0 \quad \text{and} \quad 2x-1=0 \]
\[y= 2-x \quad \text{and} \quad x= \frac{1}{2} \]
\[\Rightarrow y= 2 – \frac{1}{2} \]
\[\Rightarrow y= \frac{3}{2} \]
\(\textbf{Answer: E} \)
Question 29
For \( a, b, c \in Z^+ \),
\[\begin{aligned}a + 2b-c = &13 \\ a-b+2c =&7 \end{aligned}\] then what is the maximum value of a?
\[
\text{A)} 6 \quad
\text{B) } 7 \quad
\text{C) } 8 \quad
\text{D) } 9 \quad
\text{E) } 10
\]
Solution:
To evaluate more easily, let’s reduce the number of variables in the system of equations from three to two. To do this, let’s multiply both sides of the second equation by 2 and add the equations side by side.
\[\begin{aligned}
a + 2b-c = &13 \\
+ \quad 2a-2b+4c = &14 \\
\hline
3a+3c = 27
\end{aligned}\]
\[ 3a+3c = 27 \Rightarrow 3(a+ c )= 27 \]
\[\Rightarrow a+c =9 \]
For a to take its maximum value, c must take its minimum value. Since \(a, b \) and \(c \) are positive integers, if \(c = 1 \) is chosen, then \(a = 8 \). For these values, \(b = 3 \) is found from both equations.
\(\textbf{Answer: C} \)
Question 30
Given that \(x, y \) and \(z \) are distinct positive integers, what is the maximum value of \( y \) that satisfies the equation \(2x + 3y + 4z = 45 \)?
\[
\text{A)} 5 \quad
\text{B) } 7 \quad
\text{C) } 9 \quad
\text{D) } 11 \quad
\text{E) } 13
\]
Solution:
In the equation \(2x + 3y + 4z = 45 \), for y to take its maximum value, the sum \(2x+ 4z \) must take its minimum value. Since the coefficient of \( z \) is greater than the coefficient of \( x \), \( z \) should be chosen as small as possible. In this case,
if \( x= 4 , z = 1 \) are chosen, \[ 2x+3y+4z=45 \Rightarrow 8+3y+4=45\]
\[ \Rightarrow y = 11 \] is obtained.
\(\textbf{Answer: D} \)
Question 31
\(x, y \) and \(z \) are positive integers.
\[x^2 \;+\; y^2 \;- \;z^2 \;+ \; 2xy = 100 \] given this, what is the value of \(z \)?
\[
\text{A)} 24 \quad
\text{B) } 22 \quad
\text{C) } 20 \quad
\text{D) } 18 \quad
\text{E) } 16
\]
Solution:
\[ x^2 \;+ \;y^2 \;- \;z^2 \;+ \;2xy\; = \; 100 \Rightarrow (x+ y )^2 \;- \; z^2 =100\] from the difference of two squares
\[\Rightarrow (x+y-z) \cdot (x+y+z) =100 \]
here, since the factors \(x+y- z \) and \( x+y+z \) are positive integers and \( x+y- z < x+y+z \)
\[ x+y- z =1, \quad x+y+z =100 \quad \text{or} \]
\[ x+y- z =2, \quad x+y+z =50 \quad \text{or} \]
\[ x+y- z =4, \quad x+y+z =25 \quad \text{or} \]
\[ x+y- z =5, \quad x+y+z =20 \quad \quad\quad \]
can be possible. However, when the resulting pairs of equations are subtracted side by side to find \(z \), it is seen that z is a positive integer only in the second pair. Accordingly,
\[\begin{aligned} x+ y+ z = &50 \\ – \quad x+y -z = &2 \\ \hline \\ 2z= 48 \Rightarrow z = 24 \end{aligned}\]
\(\textbf{Answer: A} \)
← Previous Page | Next Page →