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Find the values of J and S that make both equations true using elimination. [3K each]

① 4J+2S =0.70
② 4J+ 5S =0.85

1 Answer

2 votes

Answer:

J=0.15

S=0.05

Explanation:

4J+2S=0.70

4J+5S=0.85

Using elimination method

4J+2S=0.70 (1)

4J+5S=0.85 (2)

Subtract (1) from (2)

3S=0.15

Divide both sides by 3

3S/3=0.15/3

S=0.05

Substitute s=0.05 into (1)

4J+2S=0.70

4J+2(0.05)=0.70

4J+0.10=0.70

4J=0.70-0.10

4J=0.60

Divide both sides by 4

4J/4=0.60/4

J=0.15

CHECK:

4J+2S=0.70

4(0.15)+2(0.05)=0.70

0.60+0.10=0.70

0.70=0.70

4J+5S=0.85

4(0.15)+5(0.05)=0.85

0.60+0.25=0.85

0.85=0.85

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