# Construct 3 equations starting with x = -2.

Answer

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Hint: To get the equations starting from x = -2, multiply, divide, subtract or add common entities to both sides of x = -2, like add 5 on both sides to get the equation x + 5 = 3 and so on.

Complete step-by-step answer:

Here we have to construct 3 equations starting from x = -2. Before proceeding with this question, we must know what an equation is. In algebra, an equation can be defined as a mathematical statement consisting of an equal symbol between two algebraic expressions that have the same value. For example, 3x = 4 is an equation, in which 3x and 4 are two expressions separated by an ‘equal’ sign. An equation can have one or more variables.

Now, we take our given equation that is,

\[x=-2....\left( i \right)\]

Now, we have to construct 3 equations starting from the above equation such that the value of x remains -2.

We know that by adding common value to both RHS and LHS in the equation, the value of the variable remains constant. So, now we add 8 to both sides of equation (i), we get

\[\Rightarrow x+8=-2+8\]

\[\Rightarrow x+8=6\]

Therefore, we get the first equation as x + 8 = 6.

We know that by multiplication common values with both RHS and LHS in the equation, the value of the variable remains constant.

So, now we multiply 7 on both sides of equation (i), we get,

\[\Rightarrow 7x=7\times \left( -2 \right)\]

\[\Rightarrow 7x=-14\]

Therefore, we get the second equation as 7x = -14.

We know that by dividing common value with both RHS and LHS in the equation, the value of variables remains constant.

So, now we divide 2 with both sides of equation (i), we get,

\[\Rightarrow \dfrac{x}{2}=\dfrac{-2}{2}\]

\[\Rightarrow \dfrac{x}{2}=-1\]

Therefore, we get the third equation as \[\dfrac{x}{2}=-1\].

Therefore, we get 3 equations starting from x = -2 are,

1. \[x+8=6\]

2. \[7x=-14\]

3. \[\dfrac{x}{2}=-1\]

Note: Students must note that if we multiply, divide, subtract or add the same quantity with both LHS and RHS of the equation, then that equation remains the same. Also, after getting a new equation, students must cross-check if the value of the variable is same in the new equation or not. For example, if we take equation x + 8 = 6, by subtracting 8 from both sides, we get,

\[x+8-8=6-8\]

\[\Rightarrow x=-2\]

Hence, our equation is correct.

Complete step-by-step answer:

Here we have to construct 3 equations starting from x = -2. Before proceeding with this question, we must know what an equation is. In algebra, an equation can be defined as a mathematical statement consisting of an equal symbol between two algebraic expressions that have the same value. For example, 3x = 4 is an equation, in which 3x and 4 are two expressions separated by an ‘equal’ sign. An equation can have one or more variables.

Now, we take our given equation that is,

\[x=-2....\left( i \right)\]

Now, we have to construct 3 equations starting from the above equation such that the value of x remains -2.

We know that by adding common value to both RHS and LHS in the equation, the value of the variable remains constant. So, now we add 8 to both sides of equation (i), we get

\[\Rightarrow x+8=-2+8\]

\[\Rightarrow x+8=6\]

Therefore, we get the first equation as x + 8 = 6.

We know that by multiplication common values with both RHS and LHS in the equation, the value of the variable remains constant.

So, now we multiply 7 on both sides of equation (i), we get,

\[\Rightarrow 7x=7\times \left( -2 \right)\]

\[\Rightarrow 7x=-14\]

Therefore, we get the second equation as 7x = -14.

We know that by dividing common value with both RHS and LHS in the equation, the value of variables remains constant.

So, now we divide 2 with both sides of equation (i), we get,

\[\Rightarrow \dfrac{x}{2}=\dfrac{-2}{2}\]

\[\Rightarrow \dfrac{x}{2}=-1\]

Therefore, we get the third equation as \[\dfrac{x}{2}=-1\].

Therefore, we get 3 equations starting from x = -2 are,

1. \[x+8=6\]

2. \[7x=-14\]

3. \[\dfrac{x}{2}=-1\]

Note: Students must note that if we multiply, divide, subtract or add the same quantity with both LHS and RHS of the equation, then that equation remains the same. Also, after getting a new equation, students must cross-check if the value of the variable is same in the new equation or not. For example, if we take equation x + 8 = 6, by subtracting 8 from both sides, we get,

\[x+8-8=6-8\]

\[\Rightarrow x=-2\]

Hence, our equation is correct.

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