Respuesta :
Answer:
3. ∴ Equation is [tex]y-7=0.[/tex]
4. LHS = RHS Hence, [tex](5, 15)[/tex] is a solution of a given equation.
5. LHS ≠ RHS Hence, [tex](3, 1)[/tex] is not a solution of a given equation.
7. [tex]z=-7[/tex]
8. [tex]z=9[/tex]
9. [tex]x=1.25[/tex]
10. [tex]z=\frac{-5}{7} =-0.71428[/tex]
11. [tex]x=\frac{-15}{52} =-0.28846[/tex]
Step-by-step explanation:
3. We have given the equation [tex]3y+6=5y-8[/tex]. Now, solving the given equation we get,
⇒ [tex]5y-3y=6+8[/tex]
⇒ [tex]2y=14[/tex]
⇒ [tex]y=7[/tex]
∴ Equation is [tex]y-7=0.[/tex]
4. Given point is [tex](5, 15)[/tex], where [tex]x=5[/tex] and [tex]y=15[/tex].
Now equation is [tex]y=3x[/tex]
⇒ [tex]y-3x=0[/tex] (...[tex]i[/tex])
now putting the value of [tex]x[/tex] and [tex]y[/tex] in equation (1), we get
⇒ [tex]15-3\times5=0[/tex]
⇒ [tex]0=0[/tex] Here LHS = RHS Hence, [tex](5, 15)[/tex] is a solution of a given equation.
5. Given point is [tex](3, 1)[/tex], where [tex]x=3[/tex] and [tex]y=1[/tex].
Now equation is [tex]y=-x+3[/tex]
⇒ [tex]y+x-3=0[/tex] [tex](...ii)[/tex]
now putting the value of [tex]x[/tex] and [tex]y[/tex] in equation (11), we get
⇒ [tex]1+3-3= 0[/tex]
⇒ [tex]1\neq 0[/tex] Here LHS ≠ RHS Hence, [tex](3, 1)[/tex] is not a solution of a given equation.
7. Given equation is [tex]-4=z+3[/tex]
⇒ [tex]z=-4-3[/tex]
⇒ [tex]z=-7[/tex]
8. Given equation is [tex]z-10=-1[/tex]
⇒ [tex]z=-1+10[/tex]
⇒ [tex]z=9[/tex]
9. Given equation is [tex]2.5=2x[/tex]
⇒ [tex]x=\frac{2.5}{2}[/tex]
⇒ [tex]x=1.25[/tex]
10. Given equation is [tex]-5=z\times7[/tex]
⇒ [tex]z=\frac{-5}{7} =-0.71428[/tex]
11. Given equation is [tex]-15=52x[/tex]
⇒ [tex]x=\frac{-15}{52} =-0.28846[/tex]
