
How do you solve $8x+15=3x-20$?
Answer
560.4k+ views
Hint: It is a linear equation in one variable. There is only one unknown here so one equation is sufficient to find the value of the unknown. We will separate the constant term and variable then we can find the value of x.
Complete step by step answer:
The given equation is $8x+15=3x-20$ .
First we will bring the entire unknown to LHS and all constant to RHS, taking 3x to LHS and 15 to RHS we get
$8x-3x=-15-20$
We can write $8x-3x$ as 5x and -15-20 as -35 we can replace this in the above equation
Further solving we get
$\Rightarrow 5x=-35$
We can divide both LHS and RHS by 5 to find the value of x.
$\Rightarrow x=\dfrac{-35}{5}$
$\Rightarrow x=-7$
In this way we can find the unknown in a linear equation in one variable.
Note: Given a linear equation we saw how we can find the solution but in one condition we can not solve a linear equation in one variable, the condition is when the coefficient of the unknown in both LHS and RHS are same and the constants are not same in LHS and RHS. For example we can solve the equation $2x+5=2x+8$ actually in this condition the value of x will be tending to infinity so we can not solve for x. There is another condition where the infinite value of x is possible, the condition is when the coefficient of x is the same in both LHS and RHS and the constant is also the same in LHS and RHS. For example there are infinite possibilities for x in the equation $5x+6=5x+6$ .
Complete step by step answer:
The given equation is $8x+15=3x-20$ .
First we will bring the entire unknown to LHS and all constant to RHS, taking 3x to LHS and 15 to RHS we get
$8x-3x=-15-20$
We can write $8x-3x$ as 5x and -15-20 as -35 we can replace this in the above equation
Further solving we get
$\Rightarrow 5x=-35$
We can divide both LHS and RHS by 5 to find the value of x.
$\Rightarrow x=\dfrac{-35}{5}$
$\Rightarrow x=-7$
In this way we can find the unknown in a linear equation in one variable.
Note: Given a linear equation we saw how we can find the solution but in one condition we can not solve a linear equation in one variable, the condition is when the coefficient of the unknown in both LHS and RHS are same and the constants are not same in LHS and RHS. For example we can solve the equation $2x+5=2x+8$ actually in this condition the value of x will be tending to infinity so we can not solve for x. There is another condition where the infinite value of x is possible, the condition is when the coefficient of x is the same in both LHS and RHS and the constant is also the same in LHS and RHS. For example there are infinite possibilities for x in the equation $5x+6=5x+6$ .
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