Part a)
Answer:
As the L.H.S = R.H.S, so [tex]x = 1[/tex] is the solution to the equation [tex]3x + 4 = 7[/tex]
Step-by-step explanation:
Given the equation
[tex]3x + 4 = 7[/tex]
solving the equation for x = 1
substituting the value of x = 1 in the equation
[tex]3\left(1\right)+\:4\:=\:7[/tex]
[tex]7=7[/tex]
As the L.H.S = R.H.S, so [tex]x = 1[/tex] is the solution to the equation [tex]3x + 4 = 7[/tex]
Part b)
Answer:
As the L.H.S = R.H.S, so [tex]x = 15[/tex] is the solution to the equation [tex]\frac{x-3}{2}=6[/tex]
Step-by-step explanation:
Given the equation
[tex]\frac{x-3}{2}=6[/tex]
solving the equation for x = 15
substituting the value of x = 15 in the equation
[tex]\frac{15-3}{2}=6[/tex]
[tex]\frac{12}{2}=6[/tex]
[tex]6=6[/tex]
As the L.H.S = R.H.S, so [tex]x = 15[/tex] is the solution to the equation [tex]\frac{x-3}{2}=6[/tex]
Part c)
Answer:
As the L.H.S = R.H.S, so x = -3 in the equation is the solution to the equation [tex]x^2=-2x+3[/tex]
Step-by-step explanation:
Given the equation
[tex]x^2=-2x+3[/tex]
solving the equation for x = -3
substituting the value of x = -3 in the equation
[tex]x^2=-2x+3[/tex]
[tex]\left(-3\right)^2=-2\left(-3\right)+3[/tex]
[tex]9 = 6 + 3[/tex]
[tex]9 = 9[/tex]
As the L.H.S = R.H.S, so x = -3 in the equation is the solution to the equation [tex]x^2=-2x+3[/tex]