1001 Algebra Problems.PDF

b. 4– 3 ^2 x– 1 = ^12 – y 6 4– 3 ^2 x– 61 = 6^12 – y 2(4 – 2x) – 6 = 3(1 – y) 8 – 4x– 6 = 3 – 3y 2 – 4x= 3 – 3y 3 y= ...

d. –5[x– (3 – 4x– 5) – 5x] –2^2 = 4[2 –(x–3)] –5[x– 3 + 4x– 5) – 5x] – 4 = 4[2 – x+ 3] –5[2] – 4 = 4[5 – x] –10 – 4 = 20 – 4x – ...

a.  0 – .x 3 20 x (–0.3)(20) = –6 (Note: Remember to reverse the inequality sign when dividing both sides of an inequality b ...

c.We rewrite the given equation as an equiva- lent one solved for |x|: –3|x| + 2 = 5|x| – 14 –3|x| + 16 = 5|x| 16 = 8|x| 2 = | ...

d.Note that |a|cif and only if (acor a–c). Using this fact, we see that the values ofx that satisfy the inequality |x|3 are ...

or 1 – 4x –7. We solve these two inequalities separately: 1 – 4x 7 1 – 4x –7 –4x 6–4x –8 x – ^64 = –^32  x 2 So, the solut ...

a.Look for the selection that has a negative x- coordinate and positive y-coordinate. Since a0, it follows that –a0. The choi ...

the line using the point-slope formula y– y 1 = m(x–x 1 ), where (x 1 ,y 1 ) is the point on the line. Applying this yields the ...

the slope is m= = –^86 = –^43 .Now,we equate the expression obtained by computing the slope of this line using the points (– ...

Since the slope is ^17 , the graph of the line rises from left to right at a rate of one vertical unit up per seven horizontal ...

c.Let x= the length of the first piece. Then, 2x– 1 = the length of the second piece and 3(2x– 1) + 10 = the length of the thir ...

a.The important concept in this problem is how rate, time, and distance interrelate. It is known that distance = ratetime. We ...

b.Let x= the number of nickels in the piggy bank. Then there are 65 – xdimes in the bank. The amount contributed to the total b ...

c.The graph of the line is solid, so it is included in the solution set and the inequality describing the shaded region must in ...

20. This can be verified by choosing an arbitrary point in the shaded region, say (0,0), and observing that we can verify this ...

Set 20 (Page 51) c.Adding the two equations together yields the equation 10a= –40, the solution of which is a= –4. Now, substit ...

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 b ...

c.Multiply the second equation by 3 and add it to the first equation to obtain 14q= 98, which simplifies to q= 7. Now, substitu ...

e.Solve the second equation for ain terms ofb: b+ a= 13 a= 13 – b Substitute this expression for ain the first equation and so ...

Substitute the value ofainto the second equa- tion and solve for b: b– (–4) = 1 b+ 4 = 1 b= –3 Since a= –4 and b= –3, the value ...