CBSE Class 7 Maths Exemplar Chapter 4: Simple Equations NCERT Solutions

To find the solution for the equation $ax\; +\; b\; =\; 0$, we need to isolate the variable $x$. First, subtract $b$ from both sides of the equation: $$ax\; =\; -b$$ Next, divide both sides by $a$ (assuming $a\; \backslash neq\; 0$) to solve for $x$: $$x\; =\; \backslash frac\{-b\}\{a\}$$ Therefore, the correct option is (b) $-b/a$.

We are given the equation $ax\; =\; b$, where $a$ and $b$ are positive integers. To find the solution for $x$, we divide both sides of the equation by $a$: $$x\; =\; \backslash frac\{b\}\{a\}$$ Since both $a$ and $b$ are positive integers, their quotient $\backslash frac\{b\}\{a\}$ will also be a positive number. It might be an integer or a fraction, but it will always be positive. Therefore, the solution will always be a positive number.

The fundamental property of equations is that whatever operation you perform on one side, you must perform the same operation on the other side to maintain the equality. Adding, subtracting, and multiplying by any non-zero number are always allowed operations. However, dividing both sides by the same number is only allowed if that number is non-zero. Division by zero is undefined and therefore not allowed. Thus, dividing both sides of the equation by the same number (which could potentially be zero) is the operation that is not always allowed.

Let's solve each equation to determine the nature of its solution: (a) $2x\; +\; 6\; =\; 0$ $2x\; =\; -6$ $x\; =\; -3$ (This is an integer). (b) $3x\; -\; 5\; =\; 0$ $3x\; =\; 5$ $x\; =\; \backslash frac\{5\}\{3\}$ (This is a fraction). (c) $5x\; -\; 8\; =\; x\; +\; 4$ $5x\; -\; x\; =\; 4\; +\; 8$ $4x\; =\; 12$ $x\; =\; 3$ (This is an integer). (d) $4x\; +\; 7\; =\; x\; +\; 2$ $4x\; -\; x\; =\; 2\; -\; 7$ $3x\; =\; -5$ $x\; =\; \backslash frac\{-5\}\{3\}$ (This is a fraction). The question asks for a solution that is neither a fraction nor an integer. However, all solutions derived are either integers or fractions. Re-examining the options and the typical intent of such questions, it's likely asking for a solution that is a fraction (and not an integer). Both (b) and (d) yield fractional solutions. If the question implies a non-integer rational number, then both (b) and (d) fit. Let's assume the question meant 'a fraction that is not an integer'. In that case, both (b) and (d) are valid. However, if we must choose one, and considering the provided source answer points to (d), we will proceed with that. The solution $x\; =\; -5/3$ is a fraction and not an integer.

Let's solve each equation to see if the solution is an integer: (a) $5y\; -\; 3\; =\; -18$ $5y\; =\; -18\; +\; 3$ $5y\; =\; -15$ $y\; =\; -3$ (This is an integer). (b) $3x\; -\; 9\; =\; 0$ $3x\; =\; 9$ $x\; =\; 3$ (This is an integer). (c) $3z\; +\; 8\; =\; 3\; +\; z$ $3z\; -\; z\; =\; 3\; -\; 8$ $2z\; =\; -5$ $z\; =\; \backslash frac\{-5\}\{2\}$ (This is a fraction, not an integer). (d) $9y\; +\; 8\; =\; 4y\; -\; 7$ $9y\; -\; 4y\; =\; -7\; -\; 8$ $5y\; =\; -15$ $y\; =\; -3$ (This is an integer). The equation $3z\; +\; 8\; =\; 3\; +\; z$ has a solution $z\; =\; -5/2$, which is a fraction and not an integer. Therefore, this is the equation that cannot be solved in integers.

We are given the equation $7x\; +\; 4\; =\; 25$. To find the value of $x$, we first isolate the term containing $x$. Subtract 4 from both sides of the equation: $$7x\; =\; 25\; -\; 4$$ $$7x\; =\; 21$$ Now, divide both sides by 7 to solve for $x$: $$x\; =\; \backslash frac\{21\}\{7\}$$ $$x\; =\; 3$$ Thus, the value of $x$ is 3.

We need to solve the equation $3x\; +\; 7\; =\; -20$ for $x$. First, isolate the term with $x$ by subtracting 7 from both sides of the equation: $$3x\; =\; -20\; -\; 7$$ $$3x\; =\; -27$$ Next, divide both sides by 3 to find the value of $x$: $$x\; =\; \backslash frac\{-27\}\{3\}$$ $$x\; =\; -9$$ The solution to the equation is -9.