EQUATIONS

$3(5)+5=15+5=20$ ? — yes, $x=5$ is a solution.

$3(6)+5=18+5=23\ne20$ — no, $x=6$ is NOT a solution.

$x=8$: LHS$=2(8)-7=9$, RHS$=8+3=11$. Not equal. $x=10$: LHS$=2(10)-7=13$, RHS$=10+3=13$. Equal! $x=12$: LHS$=2(12)-7=17$, RHS$=12+3=15$. Not equal.

$x=10$

$x=3$: $3^2=9$ ?. $x=-3$: $(-3)^2=9$ ?. Both work!

$x=3$: $2(3)=6$ ?. $x=-3$: $2(-3)=-6\ne6$ ?. Only $x=3$ works.

Squaring 'hides' the sign of the original number (since both a positive and its negative give the same square), so equations involving $x^2$ can have two solutions. Linear equations (just $x$, no power) don't have this sign-hiding effect, so they generally have exactly one solution.

Guessing and checking might miss the correct solution entirely if you don't happen to try the right number, and for equations with non-integer or unusual solutions, you could guess forever without finding it — it doesn't scale well and gives no certainty you've found ALL solutions.

Algebraic methods systematically isolate the variable through valid, reversible steps, guaranteeing you find the EXACT solution(s) without guessing — and the number of solutions an equation can have is determined by its structure (e.g. linear equations have exactly one), so you know when you're done.

$4(4.50)+2=18+2=20$ ? — yes, $x=4.50$ satisfies the equation.

A negative price per item makes no real-world sense — you cannot pay a negative amount of money for a product. Even if $x=-3$ solved the equation mathematically (in this case it doesn't, but hypothetically), it would need to be rejected as an invalid answer in this real context.

$x=2$: $3(4)-5(2)-2=12-10-2=0$ ? solution. $x=-\frac{1}{3}$: $3(\frac{1}{9})-5(-\frac{1}{3})-2=\frac{1}{3}+\frac{5}{3}-2=2-2=0$ ? solution. $x=1$: $3-5-2=-4\ne0$ ? not a solution. TWO of the three proposed values ($x=2$ and $x=-\frac{1}{3}$) are genuine solutions.

$x+5=12$: $x=7$ — exactly ONE solution. $x^2=16$: $x=4$ or $x=-4$ — exactly TWO solutions. $x^2+1=0$: $x^2=-1$, which has NO real solution (since a real number squared can never be negative) — ZERO real solutions.

A linear equation like $x+5=12$ involves $x$ to the power 1 only, so solving it always isolates $x$ to a single unique value via straightforward inverse operations. A quadratic equation involves $x^2$, which introduces the possibility of TWO different values of $x$ giving the same square (like $4^2=(-4)^2=16$) — but depending on the specific numbers involved, a quadratic might have two solutions (like $x^2=16$), exactly one repeated solution (a special case), or NO real solutions at all (like $x^2=-1$, since no real number squares to a negative).

$2(5)-3=10-3=7$ ?.

$2(5)=10$ ?, confirming $x=5$ satisfies the rearranged form too. This must ALWAYS be true because a valid rearrangement only involves performing the SAME operation on both sides of the equation (maintaining balance) — this never changes which values of $x$ make the two sides equal, it only changes how the equation LOOKS, not what it fundamentally represents.

The safest, clearest approach is to substitute $x=4$ FIRST into the original equation, then simplify each side SEPARATELY and compare. Testing $x=4$: LHS $=\dfrac{2(4)+6}{4-1}=\dfrac{14}{3}$. RHS $=\dfrac{14}{3}$. Since LHS = RHS, $x=4$ IS a valid solution — substituting first and simplifying each side independently avoids the confusion of mixing algebraic rearrangement with a specific numerical test value.

$A=6(6+8)=6(14)=84$ m². Since $84\ge80$, YES, $x=6$m meets the requirement.

$x=5$: $A=5(13)=65$m² — does NOT meet 80m². Testing further: $x=5.5$: not whole number, skip. Since $x=5$ gives 65 (too small) and $x=6$ gives 84 (sufficient), the smallest WHOLE-number width meeting the requirement is $x=6$m.