Question

practice solving systems of equations numerically and graphically part 1 match each system with its solution. test the pairs of numbers to see which ones make both equations true. (a) \

$$\begin{cases} y = 3x - 7 \\\\ y = \\frac{1}{3}x + 1 \\end{cases}$$

(b) \

$$\begin{cases} 4x - 2y = 10 \\\\ 3x - y = 5 \\end{cases}$$

(c) \

$$\begin{cases} y = -2x + 7 \\\\ 2x - y = 5 \\end{cases}$$

1. (i) (0, -5) (ii) (2, 6) (iii) (3, 1) (iv) (3, 2) (v) (1, 4) (vi) (3, 10)

Explanation:

Step1: Test system (a) with pairs

Substitute each pair into

$$\begin{cases} y=3x-7 \\ y=\frac{1}{3}x+1 \end{cases}$$

:

• For (0, -5): $-5

eq 3(0)-7=-7$; $-5
eq \frac{1}{3}(0)+1=1$ → No

• For (2, 6): $6

eq 3(2)-7=-1$; $6
eq \frac{1}{3}(2)+1=\frac{5}{3}$ → No

• For (3, 1): $1=3(3)-7=2$? No; $1=\frac{1}{3}(3)+1=2$? No → No
• For (3, 2): $2=3(3)-7=2$; $2=\frac{1}{3}(3)+1=2$ → Yes
• For (1, 4): $4

eq 3(1)-7=-4$; $4
eq \frac{1}{3}(1)+1=\frac{4}{3}$ → No

• For (3, 17): $17

eq 3(3)-7=2$; $17
eq \frac{1}{3}(3)+1=2$ → No

Step2: Test system (b) with pairs

Substitute each pair into

$$\begin{cases} 4x-2y=10 \\ 3x-y=5 \end{cases}$$

:

• For (0, -5): $4(0)-2(-5)=10=10$; $3(0)-(-5)=5=5$ → Yes
• For (2, 6): $4(2)-2(6)=8-12=-4

eq 10$; $3(2)-6=0
eq 5$ → No

• For (3, 1): $4(3)-2(1)=12-2=10=10$; $3(3)-1=8

eq 5$ → No

• For (3, 2): $4(3)-2(2)=12-4=8

eq 10$; $3(3)-2=7
eq 5$ → No

• For (1, 4): $4(1)-2(4)=4-8=-4

eq 10$; $3(1)-4=-1
eq 5$ → No

• For (3, 17): $4(3)-2(17)=12-34=-22

eq 10$; $3(3)-17=9-17=-8
eq 5$ → No

Step3: Test system (c) with pairs

Substitute each pair into

$$\begin{cases} y=-2x+7 \\ 2x-y=5 \end{cases}$$

:

• For (0, -5): $-5

eq -2(0)+7=7$; $2(0)-(-5)=5=5$ → No

• For (2, 6): $6

eq -2(2)+7=3$; $2(2)-6=4-6=-2
eq 5$ → No

• For (3, 1): $1=-2(3)+7=1$; $2(3)-1=6-1=5=5$ → Yes
• For (3, 2): $2

eq -2(3)+7=1$; $2(3)-2=4
eq 5$ → No

• For (1, 4): $4=-2(1)+7=5$? No; $2(1)-4=-2

eq 5$ → No

• For (3, 17): $17

eq -2(3)+7=1$; $2(3)-17=6-17=-11
eq 5$ → No

Answer:

(a) ↔ iv. (3, 2)
(b) ↔ i. (0, -5)
(c) ↔ iii. (3, 1)