Finally, substitute this into the second equa-
tion and solve for a:
6 + 2a= –4
2 a= –10
a = –5
- c.Multiply the first equation by 8 to obtain
the equivalent equation 4x+ 48y= 56. Adding
this to the second equation in the system results
in the equation 33y= 66, which simplifies to
y= 2.
- d.Divide the second equation by 2 and add it
to the first equation to obtain the equation 7a=
21, the solution of which is a= 3. Now, substi-
tute the value ofainto the first equation and
solve for b:
4(3) + 6b= 24
12 + 6b= 24
6 b= 12
b= 2
Since a= 3 and b= 2, the value ofa+ b=
3 + 2 = 5.
- b.First, simplify the first equation by multi-
plying (a+ 3) by ^12 to obtain the equivalent
equation ^12 a+ ^32 – b= –6. Then, subtract ^32
from both sides to further obtain ^12 a– b= –^125 .
Next, multiply the equation by –6 and
add it to the second equation to obtain the
equation 4b= 40, or b= 10. Now, substitute
the value ofbinto the second equation and
solve for a:
3 a– 2(10) = –5
3 a– 20 = –5
3 a= 15
a= 5
Since a= 5 and b= 10, the value ofa+ b= 5 +
10 = 15.
- b.Multiply the first equation to 5 and simplify
to obtain c– d= 10. Then, subtract the second
equation from this to obtain 5d= 10, or d= 2.
Now, substitute this value into the second
equation and solve for c:
c– 6(2) = 0
c = 12
So, the value ofdcis ^122 = 6.
- a.Multiply the second equation by 2, then add
to the first equation to obtain –7x= –63, which
simplifies x= 9. Now, substitute this value into
the second equation and solve for y:
–9 – y= –6
So, the value ofxyis (9)(–3)= –27.
- e.First, simplify the first equation by multi-
plying (x– 1) by 9 to obtain 9(x– 1) = 9x– 9.
Then, add 9 and 4yto both sides of the equa-
tion. The first equation becomes 9x+ 4y= 11.
Multiply the second equation by –2 and add it
to the first equation to obtain –5x= 5 or x= –1.
Now, substitute the value ofxinto the second
equation and solve for y:
2 y+ 7(–1) = 3
2 y– 7 = 3
2 y= 10
y= 5
Since y= 5 and x= –1, the value of (y– x)^2 =
(5 – (–1))^2 = 6^2 = 36.
ANSWERS & EXPLANATIONS–
