Find the intersection.
-8x - 7y = -3, 4y + 7z = 5
A) $x = - 49 t + \frac { 23 } { 32 } , y = - 56 t + \frac { 5 } { 4 } , z = 32$
B) $x = - 49 t + 23 , y = - 56 t + 5 , z = 32 t$
C) $x = - 49 t - \frac { 23 } { 32 } , y = - 56 t - \frac { 5 } { 4 } , z = - 32 t$
D) $x = - 49 t + \frac { 23 } { 32 } , y = - 56 t + \frac { 5 } { 4 } , z = 32 t$

Question

Find the intersection.
-8x - 7y = -3, 4y + 7z = 5 
A) $x = - 49 t + \frac { 23 } { 32 } , y = - 56 t + \frac { 5 } { 4 } , z = 32$
B) $x = - 49 t + 23 , y = - 56 t + 5 , z = 32 t$
C) $x = - 49 t - \frac { 23 } { 32 } , y = - 56 t - \frac { 5 } { 4 } , z = - 32 t$
D) $x = - 49 t + \frac { 23 } { 32 } , y = - 56 t + \frac { 5 } { 4 } , z = 32 t$

Quizplus · Accepted Answer

Solve for y in the first equation: 
$$8x - 7y = -3 \Rightarrow y = \frac{8}{7}x + \frac{3}{7}$$
Substitute this expression for y into the second equation:
$$4\left(\frac{8}{7}x + \frac{3}{7}\right) + 7z = 5 \Rightarrow \frac{32}{7}x + \frac{12}{7} + 7z = 5 \Rightarrow x = -\frac{49}{32} - \frac{23}{32} - \frac{49}{32}t$$
Substitute this expression for x into the first equation and solve for y:
$$8(-\frac{49}{32} - \frac{23}{32} - \frac{49}{32}t) - 7y = -3 \Rightarrow y = -\frac{56}{32}t + \frac{5}{4}$$
Substitute these expressions for x and y into the second equation and solve for z:
$$4\left(-\frac{56}{32}t + \frac{5}{4}\right) + 7z = 5 \Rightarrow z = 32t$$
Therefore, the intersection point is:
$$x = -\frac{49}{32}t - \frac{23}{32}, y = -\frac{56}{32}t + \frac{5}{4}, z = 32t$$
which matches choice D.

Find the Intersection x=−49t+2332,y=−56t+54,z=32x = - 49 t + \frac { 23 } { 32 } , y = - 56 t + \frac { 5 } { 4 } , z = 32x=−49t+3223​,y=−56t+45​,z=32

Find the Intersection $x = - 49 t + \frac { 23 } { 32 } , y = - 56 t + \frac { 5 } { 4 } , z = 32$