Mathematics
Grade-8
Easy

Question

Number of solutions for 5 over 2 left parenthesis x minus 5 right parenthesis equals fraction numerator negative 2 over denominator 5 end fraction left parenthesis x minus 4 right parenthesis is

  1. One solution
  2. No solution
  3. Infinity many solution
  4. Cannot be determined

hintHint:

An equation contains constants, variables and an equal sign. It has two sides called LHS and RHS. A linear equation in one variable is an equation with only one variable and the highest power of the variable should be one. Here, we have to find the number of solutions of the given linear equation in one variable which can be done by writing the given equation in standard form and comparing the it with the standard equation.

The correct answer is: One solution


The given equation is 5 over 2 left parenthesis x minus 5 right parenthesis equals fraction numerator negative 2 over denominator 5 end fraction left parenthesis x minus 4 right parenthesis
Here, we have to find the number of solutions for the given equation.
Step 1: Solve the bracket.
fraction numerator 5 x over denominator 2 end fraction minus 25 over 2 equals fraction numerator negative 2 x over denominator 5 end fraction plus 8 over 5
Since, the variables and constants in both sides are different, so this equation don't have infinitely many solutions.
Bring all the terms containing variables in one side by operating same numbers on both sides to balance the equation.
Step 2: Multiply 10 on both sides.(LCM of 2 and 5 is 10.)
rightwards double arrow open parentheses fraction numerator 5 x over denominator 2 end fraction minus 25 over 2 close parentheses cross times 10 equals space open parentheses fraction numerator negative 2 x over denominator 5 end fraction plus 8 over 5 close parentheses cross times 10
rightwards double arrow 5 left parenthesis 5 x minus 25 right parenthesis equals space 2 left parenthesis negative 2 x plus 8 right parenthesis
Step 3: Solve the brackets on both sides.
rightwards double arrow25x-125= -4x+16
Step 4: Add 4x on both sides.
rightwards double arrow25x-125+4x= -4x+16+4x
rightwards double arrow25x-125+4x= 16
Step 5: Add 125 on both sides.
rightwards double arrow25x-125+4x+125= 16+125
rightwards double arrow25x+4x= 16+125
Step 6: Solve both sides.
rightwards double arrow29x= 141      ...........(1)
Step 7: Write the eq. 1 in standard form,i.e., ax+b=0
rightwards double arrow29x-141=0       ...............(2)
Step 8: Compare eq. 2 with the standard equation.
On comparing eq. 2 with the standard equation we have, a= 29 and b= -141.
Since, x= fraction numerator negative b over denominator a space end fraction w h e r e comma space a not equal to 0. 
So, the equation has only one solution.
therefore, the correct option is a,i.e., one solution.

Standard form of linear equation in one variable is ax+b=0. If x=fraction numerator negative b over denominator a space end fraction w h e r e comma space a not equal to 0, then the solutions will be one. If both sides of the equation have same variable and constants then the equation has infinitely many solution.

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