Question

Show that each of the following values for the given variable is a solution to the

Answer 1

                                                      Part a)

Answer:

As the L.H.S = R.H.S, so $x = 1$ is the solution to the equation $3x + 4 = 7$

Step-by-step explanation:

Given the equation

$3x + 4 = 7$

solving the equation for x = 1

substituting the value of x = 1 in the equation

$3\left(1\right)+\:4\:=\:7$

$7=7$

As the L.H.S = R.H.S, so $x = 1$ is the solution to the equation $3x + 4 = 7$

                                                     Part b)

Answer:

As the L.H.S = R.H.S, so $x = 15$ is the solution to the equation $\frac{x-3}{2}=6$

Step-by-step explanation:

Given the equation

$\frac{x-3}{2}=6$

solving the equation for x = 15

substituting the value of x = 15 in the equation

$\frac{15-3}{2}=6$

$\frac{12}{2}=6$

$6=6$

As the L.H.S = R.H.S, so $x = 15$ is the solution to the equation $\frac{x-3}{2}=6$

                                                   Part c)

Answer:

As the L.H.S = R.H.S, so x = -3 in the equation is the solution to the equation $x^2=-2x+3$

Step-by-step explanation:

Given the equation

$x^2=-2x+3$

solving the equation for x = -3

substituting the value of x = -3 in the equation

$x^2=-2x+3$

$\left(-3\right)^2=-2\left(-3\right)+3$

$9 = 6 + 3$

$9 = 9$

As the L.H.S = R.H.S, so x = -3 in the equation is the solution to the equation $x^2=-2x+3$