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Equation - Linear Equation

For COMPETITION
Number of Total Problems: 4.
FOR PRINT ::: (Book)

Problem Num : 1

From : AMC10B

Type:

Section:Equation

Theme:
Adjustment# : 0

Difficulty: 1

'
The lines $x=frac{1}{4}y+a$ and $y=frac{1}{4}x+b$ intersect at the point $(1,2)$ . What is $a+b$ ?
$mathrm{(A) } 0qquad mathrm{(B) } frac{3}{4}qquad mathrm{(C) } 1qquad mathrm{(D) } 2qquad mathrm{(E) } fr...$
'

Category Linear Equation

Analysis

Solution/Answer
Since $(1,2)$ is a solution to both equations, plugging in $x=1$ and $y=2$ will give the values of $a$ and $b$ .
$1 = frac{1}{4} cdot 2 + a$
$a = frac{1}{2}$
$2 = frac{1}{4} cdot 1 + b$
$b = frac{7}{4}$
So: $a+b = frac{1}{2} + frac{7}{4} = frac{9}{4} Rightarrow E$

Answer:

Problem Num : 2

From : AMC10

Type:

Section:Equation

Theme:
Adjustment# : 0

Difficulty: 1

'
The equations $2x + 7 = 3$ and $bx - 10 = - 2$ have the same solution. What is the value of $b$ ?
$(mathrm {A}) -8 qquad (mathrm {B}) -4 qquad (mathrm {C}) 2 qquad (mathrm {D}) 4 qquad (mathrm {E}) 8$
'

Category Linear Equation

Analysis

Solution/Answer
$2x + 7 = 3 Longrightarrow x = -2, quad -2b - 10 = -2 Longrightarrow -2b = 8 Longrightarrow b = -4 mathrm{(B)}$

Answer:

Problem Num : 3

From : AMC10

Type:

Section:Equation

Theme:
Adjustment# : 0

Difficulty: 1

'
The larger of two consecutive odd integers is three times the smaller. What is their sum?
$ext{(A)} 4 qquad ext{(B)} 8 qquad ext{(C)} 12 qquad ext{(D)} 16 qquad ext{(E)} 20$
'

Category Linear Equation

Analysis

Solution/Answer
Let the two consecutive odd integers be $a$ , $a+2$ . Then $a+2 = 3a$ , so $a = 1$ and their sum is $4 mathrm{(A)}$ .

Answer:

Problem Num : 4

From : AMC10B

Type:

Section:Equation

Theme:
Adjustment# : 0

Difficulty: 1

'
What is the sum of all the solutions of $x = left|2x-|60-2x| ight|$ ?
$extbf{(A)} 32 qquad extbf{(B)} 60 qquad extbf{(C)} 92 qquad extbf{(D)} 120 qquad extbf{(E)} 124$
'

Category Linear Equation

Analysis

Solution/Answer
We evaluate this in cases:
Case 1 $x<30$
When $x<30$ we are going to have $60-2x>0$ . When $x>0$ we are going to have $|x|>0implies x>0$ and when $-x>0$ we are going to have $|x|>0implies -x>0$ . Therefore we have $x=|2x-(60-2x)|$ . $x=|2x-60+2x|implies x=|4x-60|$
Subcase 1 $30>x>15$
When $30>x>15$ we are going to have $4x-60>0$ . When this happens, we can express $|4x-60|$ as $4x-60$ . Therefore we get $x=4x-60implies -3x=-60implies x=20$ . We check if $x=20$ is in the domain of the numbers that we put into this subcase, and it is, since $30>20>15$ . Therefore $20$ is one possible solution.
Subcase 2 $x<15$
When $x<15$ we are going to have $4x-60<0$ , therefore $|4x-60|$ can be expressed in the form $60-4x$ . We have the equation $x=60-4ximplies 5x=60implies x=12$ . Since $12$ is less than $15$ , $12$ is another possible solution. $x=|2x-|60-2x||$
Case 2 : $x>30$
When $x>30$ , $60-2x<0$ . When $x<0$ we can express this in the form $-x$ . Therefore we have $-(60-2x)=2x-60$ . This makes sure that this is positive, since we just took the negative of a negative to get a positive. Therefore we have $x=|2x-(2x-60)|$
$x=|2x-2x+60|$
$x=|60|$
$x=60$
We have now evaluated all the cases, and found the solution to be ${60,12,20}$ which have a sum of $oxed{ extbf{(C)} 92}$

Answer:

Array ( [0] => 8059 [1] => 7810 [2] => 7846 [3] => 8134 ) 4