1001 Algebra Problems.PDF

319. c.Multiply the second equation by 3 and add
     it to the first equation to obtain 14q= 98, which
     simplifies to q= 7. Now, substitute the value of
     qinto the second equation and solve for p:

–5p+ 2(7) = 24

–5p+ 14 = 24

–5p= 10

p= –2

Since p= –2 and q= 7, the value of (p+ q)^2 =

(–2 + 7)^2 = 5^2 = 25.

320. b.Multiply the first equation by 2 to obtain
     8 x– 6y= 20 and the second equation by 3 to
     obtain 15x+ 6y= 3. Then, add these equations
     to obtain 23x= 23, which simplifies to x= 1.
     Now, substitute this into the first equation and
     solve for y:

4(1) – 3y= 10

–3y = 6

y = –2

So, the solution of the system is x= 1,y= –2.

Set 21 (Page 53)

321. a.Since the first equation is already solved for
     x, substitute it directly into the second equation
     and solve for y:

2(–5y) + 2y= 16

–10y+ 2y= 16

–8y= 16

y= – 2

Now, substitute this value for yinto the first

equation to find the corresponding value ofx:

x= –5(–2) = 10. Hence, the solution of the

system is x = 10,y= –2.

322. d.Solve the first equation for yin terms ofx:

2 x+ y= 6

y= 6 – 2x

Substitute this expression for yin the second

equation and solve for x:

6– 2 ^2 x+ 4x= 12

3 – x+ 4x = 12

3 x+ 3 = 12

3 x= 9

x= 3

323. d.Solve the first equation for ain terms ofb
     by multiplying both sides of the equation by
     2 to obtain a= 2b+ 2. Now, substitute this
     expression for ain the second equation to
     find b:

3(2b+ 2 – b) = –21

3(b+ 2) = –21

3 b+ 6 = –21

3 b= –27

b= –9

Substitute the value ofbinto the first equation

and solve for a:

 2 a= –9 + 1

 2 a= –8

a= –16

Since a= –16 and b= –9, the value of

 ba=  ––^1  96 = ^19 ^6 = ^43 .

ANSWERS & EXPLANATIONS–