SAT Question 979c6ebc: Answer, Explanation & Desmos Solution

$7 x + 6 y = 5$

$28 x + 24 y = 20$

For each real number $r$ , which of the following points lies on the graph of each equation in the xy-plane for the given system?

Difficulty:Hard

$(r, - \frac{6 r}{7} + \frac{5}{7})$

$(r, \frac{7 r}{6} + \frac{5}{6})$

$(\frac{r}{4} + 5, - \frac{r}{4} + 20)$

$(- \frac{6 r}{7} + \frac{5}{7}, r)$

Explanation

Correct answer: D.

Choice D is correct. Dividing each side of the second equation in the given system by $4$ yields $7 x + 6 y = 5$ . It follows that the two equations in the given system are equivalent and any point that lies on the graph of one equation will also lie on the graph of the other equation. Substituting $r$ for $y$ in the equation $7 x + 6 y = 5$ yields $7 x + 6 r = 5$ . Subtracting $6 r$ from each side of this equation yields $7 x = - 6 r + 5$ . Dividing each side of this equation by $7$ yields $x = - \frac{6 r}{7} + \frac{5}{7}$ . Therefore, the point $(- \frac{6 r}{7} + \frac{5}{7}, r)$ lies on the graph of each equation in the xy-plane for each real number $r$ .

Desmos solution:

Note: The listed r-values make choice D satisfy both equations, showing its coordinates lie on both graphs.

Enter in Desmos:

$r_{1} = [- 2, 0, 3]$
$(- \frac{6 r _{1}}{7} + \frac{5}{7}, r_{1})$
$7 (- \frac{6 r _{1}}{7} + \frac{5}{7}) + 6 r_{1}$
$28 (- \frac{6 r _{1}}{7} + \frac{5}{7}) + 24 r_{1}$

Reviewed and rewritten by CookSAT

Why the other choices are wrong

Choice A

Choice A is incorrect. Substituting $r$ for $x$ in the equation $7 x + 6 y = 5$ yields $7 r + 6 y = 5$ . Subtracting $7 r$ from each side of this equation yields $6 y = - 7 r + 5$ . Dividing each side of this equation by $6$ yields $y = - \frac{7 r}{6} + \frac{5}{6}$ . Therefore, the point $(r, - \frac{7 r}{6} + \frac{5}{6})$ , not the point $(r, - \frac{6 r}{7} + \frac{5}{7})$ , lies on the graph of each equation.

Choice B

Choice B is incorrect. Substituting $r$ for $x$ in the equation $7 x + 6 y = 5$ yields $7 r + 6 y = 5$ . Subtracting $7 r$ from each side of this equation yields $6 y = - 7 r + 5$ . Dividing each side of this equation by $6$ yields $y = - \frac{7 r}{6} + \frac{5}{6}$ . Therefore, the point $(r, - \frac{7 r}{6} + \frac{5}{6})$ , not the point $(r, \frac{7 r}{6} + \frac{5}{6})$ , lies on the graph of each equation.

Choice C

Choice C is incorrect. Substituting $\frac{r}{4} + 5$ for $x$ in the equation $7 x + 6 y = 5$ yields $7 (\frac{r}{4} + 5) + 6 y = 5$ , or $(\frac{7 r}{4} + 35) + 6 y = 5$ . Subtracting $(\frac{7 r}{4} + 35)$ from each side of this equation yields $6 y = - \frac{7 r}{4} - 35 + 5$ , or $6 y = - \frac{7 r}{4} - 30$ . Dividing each side of this equation by $6$ yields $y = - \frac{7 r}{24} - 5$ . Therefore, the point $(\frac{r}{4} + 5, - \frac{7 r}{24} - 5)$ , not the point $(\frac{r}{4} + 5, - \frac{r}{4} + 20)$ , lies on the graph of each equation.

SAT Question 979c6ebc: Systems of Linear Equations answer and explanation with Desmos solution

Explanation

Enter in Desmos:

Why the other choices are wrong