Question

Solve the following equations and check your answers
(a) $5x-2=18$
(b) $\dfrac{1}{4}y+\dfrac{1}{2}=5$
(c) $3x+\dfrac{1}{5}=2-x$
(d) $8x+5=6x-5$

Answer 1

Hint: First solve the equation normally using algebra and then put the answer back into expression to check whether it is satisfying

Complete step-by-step answer:
First, we will solve one by one and do the verification then and there near the answer
(i) The given expression is:
$5x-2=18$
By adding 2 on both sides, we get:
$5x-2+2=18+2$
By simplifying, we get:
$5x=20$
By dividing with 5 on both sides, we get:
$x=4$
Justification for (i):
Substitute $x=4$ in the expression $5x-2=18$
5(4) – 2 = 18
Now we will take Left hand side and Right-Hand side separately
Left Hand side = 5(4) – 2 = 20 – 2
By simplifying, we get
Left Hand side = 18
We know:
Right Hand side = 18
Left Hand side = Right Hand side
So, $x=4$ is the solution for $5x-2=18$

(ii) The given expression is:
$\dfrac{1}{4}y+\dfrac{1}{2}=5$
By multiplying 4 on both sides, we get:
$y+2=\left( 5 \right).\left( 4 \right)$
By simplifying we get:
$y+2=20$
BY subtracting 2 on both sides we get:
$y+2-2=20-2$
By simplifying we get
$y=18$
Justification for (iii):
Substitute $y=18$ in the expression:
$\dfrac{1}{4}y+\dfrac{1}{2}=5$
Now we will take Left Hand Side and Right-Hand side separately.
Left Hand side = $\dfrac{18}{4}+\dfrac{1}{2}$
By taking least common multiple, we get:
Left Hand side = $\dfrac{18+2}{4}$
Left hand side = 5
We know
Right hand side = 5
$\Rightarrow $ Left Hand Side = Right Hand side
So, $y=18$ is solution for $\dfrac{y}{4}+\dfrac{1}{2}=5$
(iii) Given expression is:

$3x+\dfrac{1}{5}=2-x$
By multiplying with 5 on both sides, we get:
$15x+1=10-5x$
By adding $5x$ on both sides, we get:
$20x+1=10$
By subtracting 1 on both sides, we get:
$20x=9$
By dividing with 20 on both sides, we get:
$x=\dfrac{9}{20}$
Justification for (iii):
We will substitute $x=\dfrac{9}{20}$ in expression
$\Rightarrow 3\left( \dfrac{9}{20} \right)+\dfrac{1}{5}=2-\dfrac{9}{20}$
By taking LCM and simplifying on both sides, we get:
$\dfrac{27+4}{20}=\dfrac{31}{20}$
So, $x=\dfrac{9}{20}$ is solution of $3x+\dfrac{1}{5}=2-x$
(iv) $8x+5=6x-5$
By subtracting $6x$ on both sides, we get:
$8x+5-6x=-5$
By subtracting 5 on both sides, we get:
$2x=-10$
By dividing with 2 on both sides, we get:
$x=-5$
Justification for (iv)
Substitute $x=-5$ in expression $8x+5=6x-5$
8(-5) + 5 = 6(-5) – 5
-40 + 5 = -30 – 5
-35 = -35
So, $x=-5$ is solution for $8x+5=6x-5$
Therefore, found and verified all solutions for 4 equations

Note: While substituting back for justification be careful while solving or else we might end up with the wrong answer.