MHB Solve Inequality & Simul Eqns: Alg Workings Q (a,b,c)

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To solve the inequality 5(3x + 1) < 11x, first simplify to find x, noting that multiplying or dividing by a negative changes the inequality's direction. For the simultaneous equations 3x^2 + y^2 - 7 = 0 and y - 3x - 5 = 0, substitute y with 3x + 5 into the first equation to derive a quadratic equation for x, yielding two solutions. Finally, check which of these x values also satisfy the inequality 5(x + 1) < x. The discussion emphasizes clear algebraic workings and the importance of verifying solutions against all conditions.
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(a) Solve the inequality 5(3x + 1) &lt;11x, Show clear algebraic working. (2)
(b) Solve the simultaneous equations 3x 2 + y 2 – 7 = 0, y – 3x – 5 = 0 Show clear algebraic working. (4)
(c) Hence find the value of x for which 5(x + 1) &lt; x and 3x 2 + y 2 – 7 = 0 and y – 3x – 5 = 0
 
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assuming you mean ...

(a) $5(3x+1) \le 11x$

solve for $x$ as you would any other equation ...
though probably not necessary to know in this case, be aware that multiplying or dividing by a negative value changes the direction of the inequality

(b) $3x^2 + y^2 - 7 = 0$
$y - 3x - 5 = 0 \implies y = 3x+5$

I would substitute $(3x+5)$ for $y$ in the first equation and solve the resulting quadratic for $x$ ... note you'll get two ordered pairs that satisfy the system

(c) Decide which of the two (or maybe both) values for $x$ from part (b) satisfy the given inequality
 

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