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Image Example of how to solve two linear equation systems with two variables
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Cathy Vikulchik 2020-02-28 at 04:42

FIRST METHOD: EASIER. Find z in 1st equation 3z = 39 Divide both sides by 3 to find z z = 13 Substitute z by 13 in 2nd equation b-13 = 2 Find b by adding 13 to both sides b-13+13 = 2+13 b = 15 SECOND METHOD: DIFFICULT Isolate z in 2nd equation z = b-2 Substitute z in 1st equation to remain with a single unknown b 3×(b-2) = 39 3b-6=39 Isolate b by adding 6 to both sides 3b-6+6 = 39+6 3b = 45 Find b in the new equation by dividing both sides by 3 b = 15 Substitute b by 15 in 2nd equation 15-z=2 15=z+2 Isolate z by subtracting 2 to both sides 15-2=z+2-2 Find z z = 13 THIRD METHOD: DIFFICULT Eliminate z in both equations 1st step Multiply both sides of 2nd equation by 3 3 x (b-z)=3 x 2 We have a new equation 3b-3z=6 2nd step Add 1st side of 1st equation to 1st side of of new equation; then add 2nd side of 1st equation to 2nd side of new equation 3b-3z+3z=39+6 z disappears and b is isolated 3b=45 Divide both sides by 3 to find value of b b = 15 3rd step Substitute b by 15 in the second equation to isolate z 15-z = 2 Find value of z z = 15-2 z = 13 In all three cases, b = 15, z = 13 z × b = 13 × 15 Answer: z × b=195
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